Study the convergence of $\sum_\limits{n=1}^{\infty}\frac{(-1)^n}{(x+n)^p}$

Question

Ex: Study the convergence of the following series: $$\sum_\limits{n=1}^{\infty}\frac{(-1)^n}{(x+n)^p}$$

I tried to solve the problem by breaking it in the following way:

1) For $p>0$

$\sum_\limits{n=1}^{\infty}\frac{(-1)^n}{(x+n)^p}\leqslant\sum_\limits{n=1}^{\infty}\frac{1}{(x+n)^p}\leqslant\sum_\limits{n=1}^{\infty}\frac{1}{n^p}$

By the integral test $\sum_\limits{n=1}^{\infty}\frac{1}{n^p}$ the series converge for $p>1$.

2) For $p<0$

$\sum_\limits{n=1}^{\infty}\frac{(-1)^n}{n^p}=\sum_\limits{n=1}^{\infty}(-1)^n(x+n)^{-p}$, as $-p>0$ the sequence diverges once $\lim_{n\to\infty}(-1)^n(x+n)^{-p}=\infty$.

3) For $p=0$

$\sum_\limits{n=1}^{\infty}\frac{(-1)^n}{n^p}=\sum_\limits{n=1}^{\infty}(-1)^n$ which diverges.

Question:

Is my answer right?

Thanks in advance!

Answer 1

Since $\forall x\in\mathbb{R}$ eventually $x+n>0$, we have that

• for $p>0$ the series converges by Leibniz Test
• for $p>1$ the series converges absolutely by comparison with $\sum \frac1 {n^p}$
• for $p\le 0$ the series diverges