Find the sum of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}$

Question

Find the sum of $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}$$ where $p > 1$ is a real number, and given that $$\sum_{n=1}^{\infty} \frac{1}{n^{p}}$$ is a convergent $p$-series and has a value equal to the real number $L$.

I know how to show that the first series is convergent, but any help on finding its sum would be appreciated.

Answer 1

Observe that the series equals $$\sum_n\frac{1}{(2n+1)^p} - \sum_n \frac{1}{(2n)^p}$$ The second term equals $\frac{L}{2^p}$.

To evaluate the first term, write $$L = \sum_n \frac{1}{n^p} =\sum_n\frac{1}{(2n+1)^p} + \sum_n \frac{1}{(2n)^p}$$ so that $$\sum_n\frac{1}{(2n+1)^p} = L - \frac{L}{2^p}$$ By substituting in the first formula, you find the desired sum: $$L - \frac{L}{2^{p-1}}$$

Answer 2

We will subtract the terms in the series which are even, and add the terms that are odd. $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} = -\sum_{n=1}^{\infty} \frac{1}{(2n)^p} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^p} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)^p}-\frac{1}{(2n)^p}$$ $$=\sum_{n=1}^{\infty} \frac{(2n)^p-(2n-1)^p}{(2n)^p(2n-1)^p}.$$ This is a telescoping sum. Can you find the sum from here?

Answer 3

By adding in all the integers and subtracting all the even numbers, you get$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p}=\sum_{n=1}^\infty \frac{1}{n^p}-2\sum_{n=1}^{\infty} \frac{1}{(2n)^p}=(1-2^{1-p})L$$