1. ## Telescoping series

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

2. ## Re: Telescoping series

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

using properties of ln.

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

3. ## Re: 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

4. ## Re: Telescoping series

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.

5. ## Re: Telescoping series

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.

6. ## Re: Telescoping series

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

7. ## Re: Telescoping series

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.

8. ## Re: 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??

9. ## Re: Telescoping series

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

10. ## Re: Telescoping series

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))?

11. ## Re: Telescoping series

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.