Dont quite understand how i would solve the following problem:

Repersent the series as a telescopic series: $\displaystyle \sum ln(n(n+2)/(n+1)^2)$ n=1 to infiniti

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- Aug 15th 2012, 07:39 AMNFS1Telescoping series
Dont quite understand how i would solve the following problem:

Repersent the series as a telescopic series: $\displaystyle \sum ln(n(n+2)/(n+1)^2)$ n=1 to infiniti - Aug 15th 2012, 08:19 AMemakarovRe: Telescoping series
First, read about telescoping series. Then prove that

$\displaystyle \sum_{n=1}^N\ln\frac{n(n+2)}{(n+1)^2}=\ln1-\ln2-\ln(N-1)+\ln N$

using properties of ln.

Edit: The right-hand side should be $\displaystyle \ln1-\ln2-\ln(N+1)+\ln(N+2)$. - Aug 15th 2012, 08:57 AMNFS1Re: Telescoping series
but what is the series represented as a partial fraction? I get:

-1/(n+1)^2

substituting in n=1 and so on i get ln-1/4, ln-1/9...which surely cannot be right - Aug 15th 2012, 09:07 AMemakarovRe: Telescoping series
Hmm, I could not parse this sentence. Edit: OK. I parsed it, but I am not sure how to represent this series as a partial fraction.

What do you claim about this expression? Obviously, $\displaystyle \frac{n(n+2)}{(n+1)^2}\ne-\frac{1}{(n+1)^2}$.

I am not sure partial fractions are necessary here. Simplify the expression under the sum using the formulas for logarithm of product and quotient. - Aug 15th 2012, 10:11 AMNFS1Re: Telescoping series
- Aug 15th 2012, 10:37 AMHallsofIvyRe: Telescoping series
The point is that $\displaystyle ln\left(\frac{n(n+2)}{(n+1)^2}\right)= ln(n+2)+ ln(n)- 2ln(n+1)= \left(ln(n+2)- ln(n+1)\right)- \left(ln(n+1)- ln(n)\right)$

- Aug 15th 2012, 10:52 AMemakarovRe: Telescoping series
Yes, this is the best representation.

I was doing it as follows.

$\displaystyle \sum_{n=1}^N\ln\left(\frac{n(n+2)}{(n+1)^2}\right) =$

$\displaystyle \sum_{n=1}^N(\ln n+\ln(n+2)-2\ln(n+1))=$

$\displaystyle (\ln1+\ln3-2\ln2)+(\ln2+\ln4-2\ln3)+(\ln3+\ln5-2\ln4)+$ ... $\displaystyle +(\ln(N-1)+\ln(N+1)-2\ln N)+(\ln N+ \ln(N+2)-2\ln(N+1))=$

$\displaystyle \ln1-\ln2-\ln(N+1)+\ln(N+2)$

Attachment 24517

The horizontal axis corresponds to arguments of ln and the vertical axis corresponds to terms of the series. Blue dots (k, n) denote ln(k) in the nth term, and red dots (k, n) denote -2ln(k) in the nth term. Then three dots on the same vertical line correspond to ln(k) - 2ln(k) + ln(k) = 0, so only dots surrounded by the dashed line remain.

This all means that I was wrong about the partial sum in post #2. - Aug 27th 2012, 04:05 PMNFS1Re: Telescoping series
Ok so the series represented as telescopic series is ln1 - ln2 + ln(n+2) - ln (n+1).

How would i go about finding the sum of the telescopic series? Would i use log laws again so that the telescopic series is ln1/2 + ln(n+2)/(n+1)?? If so what would the answer to the sum be?? - Aug 27th 2012, 06:40 PMProve ItRe: Telescoping series
Now that you have what the sum is up to n, take the limit as n goes to infinity.

- Aug 28th 2012, 05:53 AMNFS1Re: Telescoping series
- Aug 28th 2012, 05:59 AMProve ItRe: Telescoping series