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# Determine whether the series is convergent or divergent by expressing $s_n$ as a telescoping sum (as in Examples 8). If it is convergent, find its sum.$\displaystyle \sum_{n = 2}^{\infty} \frac {2}{n^2 - 1}$

## Converges $\rightarrow \frac{3}{2}$

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##### Catherine R.

Missouri State University

##### Samuel H.

University of Nottingham

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Let's determine whether the Siri's converges or diverges by raiding the SN. This is recalled the end partial some In our case, since we're starting at two, this would just be a too all the way up to an So we'LL write. This thing is a telescoping some as they shone the examples in the sex book. And then, if it's conversion, will actually go ahead and find the sum. So here, before we do anything regarding the sum, let's just look at the A n here. So we have to Let's go ahead and factor that denominator. And then here we would have to do partial fractions so we would have to solve this equation here for and be We can multiply both sides by the denominator on the left and then you get a and minus one B and plus one. So here you been solved This for A and B and you end up with negative one for a one for Bea. So I'll need more room here to evaluate the sum, so I'll need to go to the next page here. R Sum is from two to infinity and then we just used a partial fraction So we have won over and minus one minus one over and plus one. This is after the partial fraction decomp then and of course, the witnesses are in value. Then we could write. This is a limit of S K. So now this is where the telescoping will happen. So if we won't worry about the limit till the very end, let's go ahead and deal with everything inside the Red Prentice's So this is just a finite sum here. So let's write this over here. So we start off by plugging in and equals two. So this is an equals to here. Then you plug in and equals three plugin and equals four. Maybe we'LL plug in one more in this direction if you plug in and equals five and we would keep adding in that direction. And then we will also plug in the last few terms like the very, very last. Sir, will you plug in K? We see this will show up. So this is for an equals K right before that is when you plug in K minus one and then even before that. So it's best to have to really see all the cancellation that will ever happen. If you want to be more sure, it's best to write as many terms as you can the first few terms and also not just the very last term K, but also a few terms before that. So here I am writing to more terms before this, and this is just so that I can hope that I can figure out how much cancelation there is. So let's look over here. We see that there's what's left over when we started canceling, we have a one that doesn't cancel with anything, and then do you see that there's a positive on? Happier. But there's never a negative one half because these air these negatives start at one over three, then one of our four one over five. That denominator is going to keep growing, so there's no negative one half, so the one in the one half will both stay in there. However, this positive one three cancels with the negative. The positive one four cancels with the negative, and this negative one six will eventually cancel with a positive one over six. Now it's called the Otherside involving the case. So here we see that this minus one over Kaye plus one does not cancel with anything because all the positive denominators they never reach Cape Plus one. They they only go up to K minus one. Similarly, one over. Okay, this term over here for the same exact reasoning. This negative will never cancel. Because if you look at the positive denominators they only go up to came on this one, however, everything under that which means we started looking at K minus one. Daniel, sir, getting your cancellation again and then came on this two would even cancel with the term corresponding to n equals K minus three. So everything else will cancel. And when we take the limit is kay goes to infinity. One over K goes to zero one of our K plus one goes zero. So we just have one plus a half, which is three over to sew the Siri's converges. And we recall there was two parts to this problem. We determined that it converges. And then if it is, it does converge. We were supposed to find the sum, and we just did that. And we have three over two, and that's our final answer.

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