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In this video, weβll learn how to find the πth partial sum of a series and determine the convergence or divergence of the series from the limit of its partial sum.
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We will primarily consider something called telescoping series, but we will also look at geometric series and how we can establish convergence or divergence of these series.
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A series is not a list of terms in a sequence, but the sum of the terms in that sequence.
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If a series has a finite number of terms, then itβs quite straightforward to add each term and evaluate the series.
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But when there are an infinite number of terms, itβs a little less straightforward.
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In fact, the series may not even have a finite sum.
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For example, the series one plus two plus three plus four plus all the way up to π and so on wonβt have a finite sum.
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Whereas the series given by a half plus a quarter plus an eighth plus a sixteenth and so on does because the πth term of this series gets ever smaller and smaller, eventually approaching zero as π approaches β.
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We define an infinite series as the sum of the terms π one, π two, π three all the way up to π π, and so on.
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And we denote it in short by the sum from π equals one to β of π sub π.
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The thing is it doesnβt always make sense to think about the sum of infinitely many terms.
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And so, we define something called a partial sum.
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A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first.
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It could be helpful to examine the behaviour of partial sums of infinite series.
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Formally, we say that given a series the sum from π equals one to β of π sub π, π sub π is the πth partial sum.
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And thatβs given by the sum from π equals one to π of π sub π.
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This also leads us onto the definition of convergence of these series.
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We say that if the limit as π approaches β of the πth partial sum is some real number π, then the series the sum of π π is called convergent.
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This real number π is called the sum of the series.
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Of course, if this is not true, we say that the series is called divergent.
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Now, often, partials sums can be computed with the sum function.
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Other times, weβll need to spot patterns or apply formulae to look for evidence of convergence of these series.
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Letβs Look at an example.
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Find the partial sum from the series the sum from π equals one to β of π to the power of one over π minus π to the power of one over π plus one.
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Is the series convergent or divergent?
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Remember, the partial sum of our series is the sum of the first π terms.
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In general, the πth partial sum of the series, the sum from π equals one to β of π sub π, is the sum from π equals one to π of π sub π.
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And thatβs π sub one plus π sub two plus π sub three all the way up to π sub π.
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We generally define this as π sub π.
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So, letβs find the πth partial sum for our series.
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We know itβs going to be π sub one plus π sub two plus π sub three all the way up to π sub π.
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But here, π sub π is π to the power of one over π minus π to the power of one over π plus one.
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π sub one is found by replacing π with the number one.
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So, we get π to the power of one over one minus π to the power of one plus one, which is simply π minus π to the power of one-half.
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Then, π sub two is π to the power of one-half minus π to the power of one over two plus one, or π to the power of one-third.
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In the same way, we can define the remaining terms as shown.
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This means π sub π, the πth partial sum, is given by π minus π to the power of one-half.
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Thatβs π sub one.
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Plus π to the power of one-half minus π to the power of one-third.
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Remember, thatβs π sub two.
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Plus π to the power of one-third minus π to the power of a quarter.
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That was π sub three.
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And we continue adding these terms until we get to π sub π, which is π to the power of one over π minus π to the power of one over π plus one.
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But if we look really carefully at our πth partial sum, we should see that some of the terms will cancel.
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We have negative π to the power of one-half plus π to the power of one-half.
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Well, thatβs zero.
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We then have negative π to the power of one-third plus π to the power of one-third, which is also zero.
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But letβs add in π sub π minus one.
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And when we do, we see that this process repeats all the way up to negative π to the power of one over π plus π to the power of one over π.
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And so, what this means is all weβre left with to describe our πth partial sum is π minus π to the power of one over π plus one.
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And so, the partial sum for our series is π minus π to the power of one over π plus one.
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The next part of this question asks us whether the series is convergent or divergent.
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Well, remember, we say that if the limit as π approaches β of the πth partial sum is equal to some real number π, then that means the series, the sum of π π, is convergent.
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So, letβs evaluate the limit as π approaches β of π sub π.
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In that case, thatβs the limit as π approaches β of π minus π to the power of one over π plus one.
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Now, actually, π itself is completely independent of π.
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And we should notice that as π grows larger, one over π plus one grows smaller.
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As π approaches β, one over π plus one approaches zero.
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So, the limit as π approaches β of the πth partial sum is π minus π to the power of zero.
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But of course, π to the power of zero β in fact, anything to the power of zero β is equal to one.
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So, our limit is equal to π minus one.
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This is indeed a real number.
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And so, we can say that our series is convergent.
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Now, notice how the partial sum of our series eventually had just a few terms after cancelling.
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This is called the method of differences.
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And the series is called a telescoping series.
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This is a series whose partial sums eventually only have a fixed number of terms after cancelling.
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Letβs have another look at a series of this form.
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Find the partial sum of the series the sum from π equals one to β of one over two π plus one times two π minus one.
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Is the series convergent or divergent?
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It might not instantly be obvious how weβre going to find the πth partial sum of this series, but π sub π here is a fraction.
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Weβre going to manipulate the expression for the πth term of our sequence by writing it in partial fraction form.
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Remember, this is a way of simplifying the fraction to the sum of two or more less complicated rational functions.
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In this case, we write one over two π plus one times two π minus one as π΄ over two π plus one plus π΅ over two π minus one.
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We want to make the expression on the right-hand side look like that on the left.
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And so, we create a common denominator.
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And we achieve this by multiplying the numerator and denominator of our first fraction by two π minus one and of our second fraction by two π plus one.
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Once weβve done this, we can simply add the numerators.
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And so, on the right-hand side, we get π΄ times two π minus one plus π΅ times two π plus one all over two π plus one times two π minus one.
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Notice now that the denominators on the right- and left-hand side of our equation are equal.
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This, in turn, means that for these two fractions to be equal, their numerators must themselves be equal.
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That is to say, one equals π΄ times two π minus one plus π΅ times two π plus one.
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And then, we need to work out the values of π΄ and π΅.
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And we have a couple of ways that we can do this.
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We could distribute the parentheses and equate coefficients on the left- and right-hand side of our equation.
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Alternatively, we can substitute the zeros of two π plus one times two π minus one into the entire equation.
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That is, let π equal one-half, see what happens, and let π equal negative one-half and see what happens.
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By letting π be equal to one-half, we get one equals π΄ times one minus one plus π΅ times one plus one.
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But of course, π΄ times one minus one is π΄ times zero, which is zero.
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And thatβs the whole purpose of substituting these zeros in; it leaves us with an equation purely in terms of one of our unknowns.
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In this case, this simplifies to one is equal to two π΅.
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And if we divide through by two, we find π΅ is equal to one-half.
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Letβs repeat this process for π is equal to negative one-half.
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Our equation becomes one equals π΄ times negative one minus one plus π΅ times negative one plus one.
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π΅ times negative one plus one is π΅ times zero, which is zero.
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And so, this time, we have an equation purely in terms of π΄.
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Its simplifies to one equals negative two π΄.
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And if we divide through by negative two, we find π΄ is equal to negative one-half.
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And so, we replace π΄ with negative one-half and π΅ with one-half.
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And we see that our πth term, one over two π plus one times two π minus one, can be written as negative one-half over two π plus one plus a half over two π minus one.
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Letβs clear some space and simplify this a little.
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We can write a half over two π minus one as one over two times two π minus one.
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Similarly, we can write negative a half over two π plus one as negative one over two times two π plus one.
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To find the πth partial sum, weβre going to list out the first few terms of our series.
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π sub one is found by replacing π with one.
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This is the first term in our series.
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We get one over two times two minus one minus one over two times two plus one.
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This simplifies to a half minus one-sixth.
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Then, π sub two is found by replacing π with two.
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And we get one over two times four minus one minus one over two times four plus one.
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Thatβs a sixth minus a tenth.
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In the same way, we find π sub three to be equal to a tenth minus fourteenth, π sub four to be equal to a fourteenth minus an eighteenth, and so on.
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And so, we find the π partial sum to be the sum of all of these all the way up to π sub π.
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So, thatβs a half minus a sixth plus a sixth minus a tenth plus a tenth minus fourteenth plus a fourteenth minus an eighteenth all the way up to one over two times two π minus one minus one over two times two π plus one.
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But now look what happens.
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Negative a sixth plus one-sixth is zero.
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Negative a tenth plus one-tenth is zero.
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Negative one fourteenth plus one fourteenth is zero.
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And this continues all the way up to one over two times two π minus one.
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And so, weβre actually only left with two terms in our πth partial sum.
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They are a half and negative one over two times two π plus one.
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We are almost done.
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But weβre going to add these fractions once again by creating a common denominator.
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This time, we achieve this simply by multiplying the numerator and denominator of our first fraction by two π plus one.
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Which gives us two π plus one minus one over two times two π plus one.
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And one minus one is zero.
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Then, we see that these twos cancel.
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And we have our expression for the πth partial sum of our series.
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Itβs π over two π plus one.
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The second part of this question asks us to establish whether the series is convergent or divergent.
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And so, we recall that if the limit as π approaches β of π sub π exists as some real number π, then the sum of π π, this series, is convergent.
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Well, in our case, π sub π is π over two π plus one.
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So, we need to find the limit as π approaches β of π over two π plus one.
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We canβt use direct substitution because if we were substitute in π equals β, we get β over β, which we know to be undefined.
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What we can do, though, is manipulate the fraction a little.
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We divide both the numerator and denominator by the highest power of π in our denominator.
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In this case, we divide by π.
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So, on the numerator, we get π over π, which is equal to one.
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And on the denominator, the first part is two π over π, which is two.
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And then, we add one over π.
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And now, we only have one term involving an π.
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Itβs this, one over π.
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Now, as π grows larger, one over π grows smaller, eventually approaching zero.
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And so, we find the limit as π approaches β of π sub π of our πth partial sum simplifies to one over two plus zero, which is just one-half.
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Since this exists as a real number, we can say that our series is convergent.
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So far, weβve looked at telescoping series, but there are a handful of series whose partial sums can be calculated using formulae.
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One of the key ones of these is the geometric series.
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Now, itβs outside of the scope of the video to prove these results, but we can quote them.
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In a geometric series, each turn is obtained from the preceding one by multiplying it by the common ratio π.
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We say that the πth partial sum of a geometric series is given by π sub π equals π times one minus π to the πth power over one minus π.
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Importantly, we say that the series is convergent if the absolute value of π is less than one and its sum is then given by π over one minus π.
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We also say that if the absolute value of the common ratio π is greater than or equal to one, the series is divergent.
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Weβll now have a look at an example of this form.
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Find the partial sum for the series the sum from π equals one to β of two times one-half to the power of π minus one.
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Is the series convergent or divergent?
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We could begin by listing out the first few terms of this series.
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The πth partial sum is the sum of the first π terms.
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So, itβs π one plus π two plus π three plus π four all the way up to π sub π, where here π sub π is two times a half to the power of π minus one.
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And this means that π sub one is found by replacing π with one.
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We get two times a half to the power of one minus one, which is two times a half to the power of zero, or just two.
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The second term, π sub two, is two times a half to the power of two minus one, which is two times one-half to the power of one, or just two times a half.
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The third term is two times a half to the power of three minus one, which is two times one-half squared.
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And we continue this process all the way up to the term two times a half to the power of π minus one.
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Now, at first glance, it might look like this is a really complicated partial sum.
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But, in fact, we have a special type of series.
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Itβs a geometric series.
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Each term is obtained from the preceding one by multiplying it by the common ratio π.
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Now, there is a formula we can quote to find the partial sum of a geometric series.
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Itβs given by π times one minus π to the πth power over one minus π.
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So, if we can define π and π from our series, weβll be able to quite easily find the πth partial sum.
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Well, we know that the first term in our series is two times a half to the power of zero.
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Well, a half to the power of zero is one.
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So, thatβs two times one.
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And we can, therefore, say that π is equal to two.
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And then, we can see that π is equal to one-half.
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Each time, our term is being multiplied by one-half.
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And so, substituting these values of π and π into the formula for the πth partial sum, and we get two times one minus a half to the πth power over one minus a half.
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Now, in fact, the denominator of this fraction, one minus a half, becomes one-half.
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And, of course, dividing by one-half is actually the same as multiplying by two.
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So, we can multiply the numerator of our expression by two.
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And weβre left with the πth partial sum.
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Itβs four times one minus a half to the πth power.
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The second part of this question asks us to decide whether this series is convergent or divergent.
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Well, in fact, we could find the limit as π approaches β of π sub π.
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Alternatively, we can quote a general result.
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We say that a geometric series with a common ratio of π is convergent if the absolute value of π is less than one.
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And itβs divergent if the absolute value of π is greater than or equal to one.
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Well, here, π is equal to one-half.
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And the absolute value of one-half is one-half, which is less than one.
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And so, in this case, we can say that this series is convergent.
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In fact, weβre also able to generalise this and say that if the series is convergent, itβs sum is given by π over one minus π.
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In this case, thatβs two over one minus one-half, which is equal to four.
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In this video, weβve learned that a partial sum of an infinite series is a sum of a finite number of consecutive terms beginning with the first.
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Given a series the sum from π equals one to β of π sub π, its πth partial sum is defined at π sub π.
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Which is π one plus π two plus π three all the way up to π sub π.
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We say that if the limit as π approaches β of π sub π is equal to some real member π, then the series is called convergent.
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This number π is called the sum of the series.
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And if this is not true, the series is called divergent.
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We looked primarily at telescoping series.
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And those are a series whose partial sums eventually only have a fixed number of terms after cancelling.
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And this cancellation process is often known as the method of differences.
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Finally, we looked at geometric series.
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And in a geometric series, each term is obtained from the preceding one by multiplying it by the common ratio π.
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The partial sum of a geometric series is given by π times one minus π to the πth power over one minus π.
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We also saw that our geometric series are convergent if the absolute value of π is less than one.
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And if the absolute value of π is greater than or equal to one, the geometric series is divergent.