Like

Report

Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using CAS to sum the series directly.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {3n^2 + 3n + 1}{(n^2 + n)^3} $

The series converges to $\lim _{n \rightarrow \infty} s_{n}=1$

You must be signed in to discuss.

let's use a computer algebra system to go ahead and rewrite this term. Here am Yeah. So here in Wolfram Alpha, my computer algebra system. If we scroll down here, we see a partial fraction decomposition that we can work with. So let's go ahead and replace this inside of the Sigma notation. So that's one over and cubed minus one over and plus one cube. This will be much easier to work with and well, actually can use the telescoping method because this is a telescoping series, as we'll see. So here we can rewrite this infinite sum is a limit we're now we're looking at a partial sum from one to k Mhm. So this in the green, including the Sigma, This is R S K R Keith, Partial sum. So using our the computer algebra system, we were able to find a convenient expression for the partial sum. So that's SK, and this is a decent looking expression. And the reason this is convenient is because we can use telescoping method. So now the telescoping method Well, keep writing that limit until the very end. Okay, goes to infinity. So now we start plugging in values of n we start at one, so we have one minus 1/2 cubed. And then we have one over to cube, minus 1/3 cube and so on. And then, on the other hand, we'll have one over K minus one cubed, minus one over K Cube. And then the very last term, When you plug in and equal scare, you have one over K cube minus one over K plus one cubed. Now, before we take that limit, we should go ahead and cancel as much as we can. We can't cancel the one, so the one should still be there in the final answer. However, to over three, we have a negative here. This will cancel with the positive. The 1/3 cubes would have canceled with the next term. And then here we could we see that we could cancel all the way until we get to K Cube. So we're able to cancel all those intermediate terms. However, you still have one more term left with K plus one cubed in the denominator and then finally take that limit. As K goes to infinity, this fraction goes to zero. So the entire sum just goes to one. So before we go log off. The second part of the question is to go ahead and check your answer using a computer algebra system. So here in the third window, I've taken the some from one to infinity, and the sum is also one. So our computer algebra system agrees with our some because we're both getting one, So that's our final answer.