Does $\sum \frac{(-1)^n}{n^{\lambda}}$ converge?

Question

For which value of $\lambda\in \mathbb{R}$ this serie converges. $\displaystyle \sum_{n\ge1} \frac{(-1)^n}{n^{\lambda}}$

Attempt :

let $u_n(\lambda):=\dfrac{(-1)^n}{n^{\lambda}}$

If $\lambda\le0, \quad u_n(\lambda)$ doesn't converge. Then $\displaystyle \sum_{n\ge1} u_n(\lambda)$ doesn't converge.

$\boxed{\lambda>1}$

If $\lambda >1 \quad \displaystyle \sum_{n\ge1} \lvert u_n(\lambda)\rvert$ converges, so $\displaystyle \sum_{n\ge1} u_n(\lambda)$ is absolutely convergent hence convergent.

$\boxed{\lambda>0}$

$\displaystyle \sum_{k=1}^n \dfrac{(-1)^k}{k^{\lambda}}=\sum_{k=0}^p \left(\dfrac{1}{(2k+2)^{\lambda}}-\dfrac{1}{(2k+1)^{\lambda}}\right) $

Let $g\colon x\to x^{-\lambda}$

According the mean value theorem, there exists $c_k\in (2k+1,2k+2)$ such that :

$(2k+2)^{-\lambda}-(2k+1)^{-\lambda}=-\lambda c_k^{-\lambda-1}$

As $c_k\underset{k\to +\infty}{\longrightarrow} 2k+1$ then $\dfrac{(-1)^k}{k^{\lambda}}\underset{+\infty}{\sim}\dfrac{-\lambda}{(2k+1)^{\lambda+1}}\underset{+\infty}{\sim}\dfrac{-\lambda}{(2k)^{\lambda+1}}$

So we conclude that if $\lambda >0, \qquad \displaystyle \sum_{k\ge1} \left\lvert \dfrac{-\lambda}{(2k)^{\lambda+1}}\right\rvert$ converges

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Then we can conclude $\forall \lambda>0, \quad \displaystyle \sum_{n\ge1} \frac{(-1)^n}{n^{\lambda}}$ converges

Is my reasoning right?, I've got some doubt...!

Answer 1

By the alterning serie test we have :

$$\mid \frac{(-1)^n}{n^\lambda}\mid = \frac{1}{n^\lambda}$$

Which is streaclty decreasing, so it converges for all $\lambda > 0$.

Answer 2

With $\lambda \le 0 $ use the divergence test and for $\lambda > 0 $ use the alternating series test to get the result.

The series is convergent for $\lambda > 0 $ and divergent otherwise.