Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent

Find the partial sum of the series βˆ‘_{𝑛 = 1} ^{∞} 1/(2𝑛 + 1)(2𝑛 βˆ’ 1). Is the series convergent or divergent?

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Video Transcript

Find the partial sum of the series the sum from 𝑛 equals one to ∞ of one over two 𝑛 plus one times two 𝑛 minus one. Is the series convergent or divergent?

It might not instantly be obvious how we’re going to find the 𝑛th partial sum of this series, but π‘Ž sub 𝑛 here is a fraction. We’re going to manipulate the expression for the 𝑛th term of our sequence by writing it in partial fraction form. Remember, this is a way of simplifying the fraction to the sum of two or more less complicated rational functions.

In this case, we write one over two 𝑛 plus one times two 𝑛 minus one as 𝐴 over two 𝑛 plus one plus 𝐡 over two 𝑛 minus one. We want to make the expression on the right-hand side look like that on the left. And so, we create a common denominator. And we achieve this by multiplying the numerator and denominator of our first fraction by two 𝑛 minus one and of our second fraction by two 𝑛 plus one. Once we’ve done this, we can simply add the numerators. And so, on the right-hand side, we get 𝐴 times two 𝑛 minus one plus 𝐡 times two 𝑛 plus one all over two 𝑛 plus one times two 𝑛 minus one.

Notice now that the denominators on the right- and left-hand side of our equation are equal. This, in turn, means that for these two fractions to be equal, their numerators must themselves be equal. That is to say, one equals 𝐴 times two 𝑛 minus one plus 𝐡 times two 𝑛 plus one. And then, we need to work out the values of 𝐴 and 𝐡. And we have a couple of ways that we can do this. We could distribute the parentheses and equate coefficients on the left- and right-hand side of our equation. Alternatively, we can substitute the zeros of two 𝑛 plus one times two 𝑛 minus one into the entire equation. That is, let 𝑛 equal one-half, see what happens, and let 𝑛 equal negative one-half and see what happens.

By letting 𝑛 be equal to one-half, we get one equals 𝐴 times one minus one plus 𝐡 times one plus one. But of course, 𝐴 times one minus one is 𝐴 times zero, which is zero. And that’s the whole purpose of substituting these zeros in; it leaves us with an equation purely in terms of one of our unknowns. In this case, this simplifies to one is equal to two 𝐡. And if we divide through by two, we find 𝐡 is equal to one-half.

Let’s repeat this process for 𝑛 is equal to negative one-half. Our equation becomes one equals 𝐴 times negative one minus one plus 𝐡 times negative one plus one. 𝐡 times negative one plus one is 𝐡 times zero, which is zero. And so, this time, we have an equation purely in terms of 𝐴. Its simplifies to one equals negative two 𝐴. And if we divide through by negative two, we find 𝐴 is equal to negative one-half. And so, we replace 𝐴 with negative one-half and 𝐡 with one-half. And we see that our 𝑛th term, one over two 𝑛 plus one times two 𝑛 minus one, can be written as negative one-half over two 𝑛 plus one plus a half over two 𝑛 minus one.

Let’s clear some space and simplify this a little. We can write a half over two 𝑛 minus one as one over two times two 𝑛 minus one. Similarly, we can write negative a half over two 𝑛 plus one as negative one over two times two 𝑛 plus one. To find the 𝑛th partial sum, we’re going to list out the first few terms of our series. π‘Ž sub one is found by replacing 𝑛 with one. This is the first term in our series. We get one over two times two minus one minus one over two times two plus one. This simplifies to a half minus one-sixth.

Then, π‘Ž sub two is found by replacing 𝑛 with two. And we get one over two times four minus one minus one over two times four plus one. That’s a sixth minus a tenth. In the same way, we find π‘Ž sub three to be equal to a tenth minus fourteenth, π‘Ž sub four to be equal to a fourteenth minus an eighteenth, and so on. And so, we find the 𝑛 partial sum to be the sum of all of these all the way up to π‘Ž sub 𝑛. So, that’s 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 𝑛 minus one minus one over two times two 𝑛 plus one.

But now look what happens. Negative a sixth plus one-sixth is zero. Negative a tenth plus one-tenth is zero. Negative one fourteenth plus one fourteenth is zero. And this continues all the way up to one over two times two 𝑛 minus one. And so, we’re actually only left with two terms in our 𝑛th partial sum. They are a half and negative one over two times two 𝑛 plus one. We are almost done. But we’re going to add these fractions once again by creating a common denominator.

This time, we achieve this simply by multiplying the numerator and denominator of our first fraction by two 𝑛 plus one. Which gives us two 𝑛 plus one minus one over two times two 𝑛 plus one. And one minus one is zero. Then, we see that these twos cancel. And we have our expression for the 𝑛th partial sum of our series. It’s 𝑛 over two 𝑛 plus one.

The second part of this question asks us to establish whether the series is convergent or divergent. And so, we recall that if the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛 exists as some real number 𝑆, then the sum of π‘Ž 𝑛, this series, is convergent. Well, in our case, 𝑆 sub 𝑛 is 𝑛 over two 𝑛 plus one. So, we need to find the limit as 𝑛 approaches ∞ of 𝑛 over two 𝑛 plus one. We can’t use direct substitution because if we were substitute in 𝑛 equals ∞, we get ∞ over ∞, which we know to be undefined.

What we can do, though, is manipulate the fraction a little. We divide both the numerator and denominator by the highest power of 𝑛 in our denominator. In this case, we divide by 𝑛. So, on the numerator, we get 𝑛 over 𝑛, which is equal to one. And on the denominator, the first part is two 𝑛 over 𝑛, which is two. And then, we add one over 𝑛.

And now, we only have one term involving an 𝑛. It’s this, one over 𝑛. Now, as 𝑛 grows larger, one over 𝑛 grows smaller, eventually approaching zero. And so, we find the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛 of our 𝑛th partial sum simplifies to one over two plus zero, which is just one-half. Since this exists as a real number, we can say that our series is convergent.

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