I need the answer of these series with respect to x. can anyone solve it, please?

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- Jan 20th 2013, 12:12 AMnasiriseries problem
I need the answer of these series with respect to x. can anyone solve it, please?

Attachment 26619 - Jan 20th 2013, 02:13 PMhollywoodRe: series problem
It reminds me of the series $\displaystyle \sin{x} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}$.

- Hollywood - Jan 20th 2013, 07:13 PMjohngRe: series problem
Let $\displaystyle f(x)=\sum^\infty_{i=0}(-1)^n{1\over (2n+1)3^n}x^{2n+1}$ -- notice your function is $\displaystyle {1\over x}f(x)$. Now find $\displaystyle f^\prime(x)$ and try to recognize this as a geometric series. Then integrate $\displaystyle {1\over x}f^\prime(x)$ to find your original series. Since this is obviously a text book problem, the answer is "nice".

- Jan 20th 2013, 11:41 PMhollywoodRe: series problem
Yeah, what johng suggests is the way to go. I don't know what I was thinking.

- Hollywood - Jan 21st 2013, 10:13 AMnasiriRe: series problem
Thank you very much. It works for this series. I found out this problem by simplifying a bigger problem in my master thesis in finance. It is not from a textbook. But, unfortunately I made a mistake while I was simplifying the problem. There is an additional term in numerator. It means that problem is:

(n+1)/(2n+1)*(-x^2/3)^n

It cannot be recognized as a geometric series by the same method. Is there any solution for my problem? - Jan 21st 2013, 10:22 AMnasiriRe: series problem
I think that if I replace n+1 with n+1/2+1/2, and by using the method of johng, I can solve this,too. Thank you. your idea help me to solve my problem.