Telescoping series

**NFS1** · August 15th 2012, 07:39 AM

Dont quite understand how i would solve the following problem:

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

**emakarov** · August 15th 2012, 08:19 AM

First, read about telescoping series. Then prove that

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

using properties of ln.

Edit: The right-hand side should be $\ln1-\ln2-\ln(N+1)+\ln(N+2)$ .

**NFS1** · August 15th 2012, 08:57 AM

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

**emakarov** · August 15th 2012, 09:07 AM

Originally Posted by NFS1

but what is the series represented as a partial fraction?

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.

Originally Posted by NFS1

I get:

-1/(n+1)^2

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

I am not sure partial fractions are necessary here. Simplify the expression under the sum using the formulas for logarithm of product and quotient.

**NFS1** · August 15th 2012, 10:11 AM

Originally Posted by emakarov

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, $\frac{n(n+2)}{(n+1)^2}\ne-\frac{1}{(n+1)^2}$ .

I am not sure partial fractions are necessary here. Simplify the expression under the sum using the formulas for logarithm of product and quotient.

Don't quite understand how to do that. I thought to express the series as telescopic series you should express the series as partial sums. If you can show a step by step solution it will be greatly appreciated.

**HallsofIvy** · August 15th 2012, 10:37 AM

The point is that $ln\left(\frac{n(n+2)}{(n+1)^2}\right)= ln(n+2)+ ln(n)- 2ln(n+1)= \left(ln(n+2)- ln(n+1)\right)- \left(ln(n+1)- ln(n)\right)$

**emakarov** · August 15th 2012, 10:52 AM

Originally Posted by HallsofIvy

The point is that $ln\left(\frac{n(n+2)}{(n+1)^2}\right)= ln(n+2)+ ln(n)- 2ln(n+1)= \left(ln(n+2)- ln(n+1)\right)- \left(ln(n+1)- ln(n)\right)$

Yes, this is the best representation.

I was doing it as follows.

$\sum_{n=1}^N\ln\left(\frac{n(n+2)}{(n+1)^2}\right) =$
$\sum_{n=1}^N(\ln n+\ln(n+2)-2\ln(n+1))=$
$(\ln1+\ln3-2\ln2)+(\ln2+\ln4-2\ln3)+(\ln3+\ln5-2\ln4)+$ ... $+(\ln(N-1)+\ln(N+1)-2\ln N)+(\ln N+ \ln(N+2)-2\ln(N+1))=$
$\ln1-\ln2-\ln(N+1)+\ln(N+2)$

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.

**NFS1** · August 27th 2012, 04:05 PM

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??

**Prove It** · August 27th 2012, 06:40 PM

Now that you have what the sum is up to n, take the limit as n goes to infinity.

**NFS1** · August 28th 2012, 05:53 AM

Originally Posted by Prove It

Now that you have what the sum is up to n, take the limit as n goes to infinity.

What is the sum?? is it: ln1 - ln2 + ln(n+2) - ln (n+1) or using log laws is it ln(1/2 + (n+2)/(n+1))?

**Prove It** · August 28th 2012, 05:59 AM

Originally Posted by NFS1

What is the sum?? is it: ln1 - ln2 + ln(n+2) - ln (n+1) or using log laws is it ln(1/2 + (n+2)/(n+1))?

The first is correct, but the second is not. It should be $\displaystyle \begin{align*} \ln{\left[ \frac{1(n+2)}{2(n+1)} \right]} \end{align*}$ , and now take the limit.

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