Question

Find the limit of these partial sums.

Answer 1

I think this is an arithmetic sequence.

Answer 2

It isn't arithmetic or geometric, but it doesn't converge because it's less that \(1/n^2\).

Answer 3

It's more of a harmonic sequence

Answer 4

Why does it converge when it is less than 1/n^2? Could you explain further on harmonic sequence please? :)

Answer 5

My guess is that it is telescoping.

Answer 6

How do you guess that it is telescoping?

Answer 7

So you have something like \[
\frac{2}{n(n+2)} = \frac{a}{n} + \frac{b}{n+2}
\]

Answer 8

ok..

Answer 9

\[
\frac{a}{n}+\frac{b}{n+2} = \frac{a(n+2) + b(n)}{n(n+2)} = \frac{2a + (a+b)n}{n(n+2)}
\]

Answer 10

So we have \[
2=\color{blue}{2}+\color{red}{0}n= \color{blue}{2a} + \color{red}{(a+b)}n
\]This gives us the equations: \[
2a=2\\
(a+b) = 0
\]Solving this system gives: \[
a = 1\\
b=-a=-1
\]

Answer 11

Which would imply that: \[
\frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2}
\]

Answer 12

okkk.

Answer 13

Okay, for you earlier question. We know that \(1/n^2\) converges in general, and we know that any sequence bounded by a converging sequences will converge. It's a bit like squeeze theorem.

Answer 14

OHOHH I see!

Answer 15

Okay, how do we continue this problem?

Answer 16

Well, the thing about a telescoping sequence is that it cancels itself out nicely.
Just take a look:\[
\sum_{n=4}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)=
\sum_{n=4}^{\infty}\frac{1}{n}-\sum_{n=4}^{\infty}\frac{1}{n+2}
\]Now we pull out the first two terms for the \(1/n\) series: \[
\frac{1}{4}+\frac{1}{5}+\sum_{n=6}^{\infty}\frac{1}{n}-\sum_{n=4}^{\infty}\frac{1}{n+2}
\]Next we reparametrize it. We say \(n' = n+2\). So \(n=6\implies n'=4\): \[
\frac{1}{4}+\frac{1}{5}+\sum_{n'=4}^{\infty}\frac{1}{n'+2}-\sum_{n=4}^{\infty}\frac{1}{n+2}
\]Now those two sequences are equivalent. The \(n\) and \(n'\) are dummy variables.
This means we have: \[
\sum_{n=4}^{\infty}\frac{2}{n(n+2)} = \frac{1}{4}+\frac{1}{5} =\frac{9}{20}
\]

Answer 17

How would you define a telescoping sequence? And thank you so much

Answer 18

As for why I thought it would be telescoping...

Well there are two reasons, first is that there is no obvious sum for this series. Second is that it looked like a partial fractions question.

Answer 19

I guess I would say sequence \(a_n\) is telescoping if it can be written in the form: \[
a_n = b_n - b_{n+k}
\]Once you get to the \(k\)th term, you see that it will cancel itself out.

Answer 20

\[
(b_n\cancel{-b_{n+k}})+
(b_{n+1}-b_{n+k+1})+
\ldots+

(b_{n+k}-b_{n+k2})\\

(b_n-b_{n+k})+
(b_{n+1}-b_{n+k+1})+
\ldots+

(\cancel{b_{n+k}}-b_{n+k2})
\]The \(-b_{n+k}\) and \(b_{n+k}\) cancel, this continues forever if we have an infinite series. In that case the only terms we have to worry about are: \[
b_{n} + b_{n+1}+\ldots + b_{n+k-1}
\]Since these terms do not get canceled.