1. ## alternating series

Hi,
Can somebody please tell me how to find the sum of the following alternating series?

\sum_{n=0}^{\infty} (-1)^n exp(-5*n)/(2*n+1)

Thanks,

2. ## Re: alternating series

$\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{e^{-5n}}{2n+1}$

3. ## Re: alternating series

Originally Posted by rvoy
Can somebody please tell me how to find the sum of the following alternating series?
\sum_{n=0}^{\infty} (-1)^n exp(-5*n)/(2*n+1)

Is this your question $\displaystyle \sum\limits_{n =1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{e^{5n}}\left( {2n + 1} \right)}}}~?$

If so, can you show that $\displaystyle \left( {\frac{1}{{{e^{5n}}\left( {2n + 1} \right)}}} \right)$ is a monotone decreasing null sequence?

4. ## Re: alternating series

Hi Plato,
Yes this is the series,

I know this series is absolutely convergent as it is bounded above by a geometric series. But, my question is how to find the sum
of the given alternating series.

5. ## Re: alternating series

If you take the integral of the geometric series

$\displaystyle \sum_{n=0}^\infty x^{2n}\,dx$

term-by-term, you get

$\displaystyle \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$

That gives you the 2n+1 in the denominator. I think that by setting x to a strategic value and multiplying by constants inside and outside of the sum, you'll be able to come up with the answer.

- Hollywood