1. ## series problem

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

2. ## Re: series problem

It reminds me of the series $\displaystyle \sin{x} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}$.

- Hollywood

3. ## Re: 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".

4. ## Re: series problem

Yeah, what johng suggests is the way to go. I don't know what I was thinking.

- Hollywood

5. ## Re: 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?

6. ## Re: 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.