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JH
Numerade Educator

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Problem 48 Hard Difficulty

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 {1}{n^3 - n} $

Answer

$\sum_{n=2}^{\infty} \frac{1}{n^{3}-n}=\frac{1}{4}$

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Video Transcript

Let's determine whether the Siri's convergence or diverges by expressing the end partials on. So remember the definition here to some in this case, I should get started. Two, not one. Why is that? Because the sun starts at any pose too. So this is the sum from a two a three all the way off K n where Anne is given by this expression over here. Now, before we do that, let's just go ahead and rewrite this by factoring the denominator. Come on. I was two and squared minus one and then we could factor that even further and minus one and plus one. Now what I'LL do here is let me just go ahead and I'm gonna do a trick. We'LL see in a moment why I'm doing this. Let me just multiply it by two and then make up for that by dividing by two. Now, I can also go a step further here. So this is one half the sum from an equals two to infinity. Now let me rewrite this to as and plus one minus and minus one the same terms that you see the denominator. So this n plus one and minus one in the numerator didn't come out of anywhere. It came specifically because we see these terms in the denominator and it just works that way. The denominator I'Ll just keep it as it is now we can go ahead and split this into two fractions. We're almost there were already almost ready to telescope. So that's the first fraction. And I see that end plus ones will cancel. This is precisely why we wanted those terms in the numerator we wanted to be able to cancel. Then I have my itis. So here I should have factoring out the entire sum and and this term as well will also get cancellation at minus ones will cancel. So cancel those Now let's go on to the next page. So we have one half. There's some from two no infinity. Now, on the inside we have one and minus one end, minus one over in and plus one. And now we're ready to telescope so we can write this as a limit. Kay goes to infinity. Now write the sum is a partial something And then let's still write this term in there as well. Now the next step we're ready to get to work here. Well, go ahead and evaluate this partial sun by writing else in terms so the limit will not come into play until the very end. So just keep writing that limit now, ready to go ahead and plug in some value. So here you start up by plugging into, so plug that in one times two minus one over two times three, that's the first term, then Plugin and equals three. Then one more term here, four times five. Then we will keep adding until we get to the last few terms. So here I may not have enough room here. So the very last term, when you plug in K came minus one times K minus one over K K plus one. Before that, you would have plugged in K minus one and before that. Okay, so now we start cancelling as much as we can. We see that we have cancellation here, basically left and right cancellation. And then what will What will stay this way? OK, and K plus one will stay cancellation again. And it looks like we just have that on ly The first term in the last term stays So let's go ahead and write that on the next page we have the limit. Kay goes to infinity one half and then we had our first. Her much was one half, and then we had the last term. Now you let Kate go to infinity. So this is just one half times one half, because this term here goes to zero in the limit and we get one over four. So the Siri's conversions, we telescoped and it converse toe one over four, and that's our final answer.