Convergence uniform of $\sum _{n=1}^{\infty }\:\frac{(-1)^n}{x^2+n}$

Question

I need to find $x$ that $\sum _{n=1}^{\infty }\frac{(-1)^n}{x^2+n}$ is convergence and uniform convergence

for $\require{cancel}\xcancel{x\ge0}$${\color{red} {x \in \mathbb{R}}}$ we get that $\frac{1}{\textbf{x^2}+n}$ in monotone decreasing so its Leibniz and its convergence .

now I need to check if in interval $x$ so that $\sum _{n=1}^{\infty }\frac{(-1)^n}{x^2+n}$ is uniform convergence :

$$|\sum _{k=n}^{\infty }\:\frac{(-1)^k}{x^2+k}|=|r_n(x)|\le a_{n+1}=\frac{1}{x^2+n+1}$$

$$sup_{x\in \mathbb{R}}|r_n(x)|=\frac{1}{n+1}:n\to \infty = 0$$

Is this correct ? if not how can I find a different $x$?

thanks

Answer 1

There are some errors in what you did. First you write: “for $x\ge0$ we get that $\frac 1{x^ 2+n}$ in monotone decreasing”. Why for $x\geqslant0$? What's different if $x<0$?

You can also prove that your series converges uniformly in $\mathbb R$ using Dirichlet's test for uniform convergence.