## anonymous 5 years ago use ${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$ to sum the series $Sum_{n} = {2\over{1\times3}} - {4\over{3\times5}} + ... + {(-1)^{n-1} 2^n \over{(2n-1)(2n+1)}}$

1. anonymous

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

2. anonymous

what is a good form

3. anonymous

that is exactly as the question was written

4. anonymous

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 $${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$$

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

yes that is a typo

9. anonymous

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

10. anonymous

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

11. anonymous

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. $\frac{2(n-1)}{(2(n-1) -1)(2(n-1)+1)} - \frac{2n}{(2n-1)(2n+1)}$ $\text{Using } {{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$ $\frac{1}{2(2n-3)} + \frac{1}{2(2n-1)} - \frac{1}{2(2n-1)} + \frac{1}{2(2n+1)}$

12. anonymous

I hope my logic is right.

13. anonymous

cool

14. anonymous

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.

15. anonymous

thanks for pointing in right direction

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

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 $$\huge \frac 12+ \frac{ (-1)^{(k-1)}}{2k+1}$$