use $${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$$ to sum the series
$$Sum_{n} = {2\over{1\times3}} - {4\over{3\times5}} + ... + {(-1)^{n-1} 2^n \over{(2n-1)(2n+1)}}$$

Question

use $${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$$  to sum the series
$$Sum_{n} = {2\over{1\times3}} - {4\over{3\times5}} + ... + {(-1)^{n-1} 2^n \over{(2n-1)(2n+1)}}$$

anonymous · Answer

tried allsorts here but not got anywhere, difference method etc, there must be something i'm missing

anonymous · Answer

what is a good form

anonymous · Answer

that is exactly as the question was written

anonymous · Answer

Hmm I am nowhere near the solution but have you tried solving (n-1)th term and nth term together, I think you should try solving them maybe some good pattern emerges out, I am going to try it actually but the nth term is a little complicated and I might commit some mistakes. 

Note. Using this ${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$

anonymous · Answer

i wondered with there was a rearrangement that would give me a way to use difference method

anonymous · Answer

argh I am getting nowhere, If you want you can post this problem on "Meta-math" (Group on OpenStudy).

anonymous · Answer

BTW could the $nth$ term be $\large {(-1)^{n-1} 2n \over{(2n-1)(2n+1)}}$?

anonymous · Answer

yes that is a typo

anonymous · Answer

$2n$ instead of $2^n$ can make the problem a lot simpler.

anonymous · Answer

Oh, so it is $2n$. Thanks, maybe now I can get it...

anonymous · Answer

Okay, so things get much simpler now. 
I am gonna show $nth$ and $(n-1)th$ only, we will see that every term cancels out except two terms in the whole expression i.e the last one and the first one. 

Let's assume the $nth$ term to be negative, that makes $(n-1)th$ term positive. 

$$\frac{2(n-1)}{(2(n-1) -1)(2(n-1)+1)} - \frac{2n}{(2n-1)(2n+1)}$$

$$\text{Using } {{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$$

$$\frac{1}{2(2n-3)} + \frac{1}{2(2n-1)} - \frac{1}{2(2n-1)} + \frac{1}{2(2n+1)}$$

anonymous · Answer

I hope my logic is right.

anonymous · Answer

cool

anonymous · Answer

A similar thing must happen in the $(n-2)th$ term, what I mean is the $(n-2)th$ term must have a term of $1\over 2(2n-3)$ which cancels out $1\over 2(2n-3)$ that off $(n-1)th$ term. 
This whole process must continue until the first term (first term of the expanded form of first term of the series) i.e I think $1\over 2$ 

Hmm final answer should look something like $\frac{1}{2} + \frac{1}{2(2n-1)$ or $\frac{1}{2} - \frac{1}{2(2n-1)$.

My answer might be wrong but I think the concept is right.

anonymous · Answer

thanks for pointing in right direction

anonymous · Answer

uhhh, terrible I hate it when my LaTeX goes wrong. 

$$\frac{1}{2} \pm \frac{1}{2n+1}$$

You're Welcome.

anonymous · Answer

It was a terrible typo which make the problem much harder, so now you have fixed the typo, I got the answer $ \huge  \frac 12+ \frac{ (-1)^{(n-1)}}{2k+1} $

anonymous · Answer

apologies, for that i'm not good with equation editing yet, i'm sure i'll get better, and its hard to edit once posted

anonymous · Answer

Oh it's okay, the lack of edit feature is really a bane while asking/answering this kind of  questions.

PS: *EDIT* The correct sum is $ \huge  \frac 12+ \frac{ (-1)^{(k-1)}}{2k+1} $