Uniform or absolute convergence of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^2+n^2}{n^3}$ in interval $[-1,1]$

Question

Uniform or absolute convergence of

$\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^2+n^2}{n^3}$

solution i tried-The given sequence is $\displaystyle(-1)^{n-1}\frac{x^2+n^2}{n^3}$

lets take $b_n=\displaystyle\frac{1}{n }\;\;so \;\;\frac{a_n}{b_n}$ $$\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{\frac{x^2+n^2}{n^3}}{\frac{1}{n}}=\frac {n(x^2+n^2)}{n^3}$$ $$=n^3\frac{(\frac{x^2}{n^2}+1)}{n^3}$$ $$\lim_{n \to \infty} (\frac{x^2}{n^2}+1)=1 $$

which is finite so given series is not absolutly convergent.because $b_n=\frac{1}{n}$

But i am not getting idea about unifrom convergence, i tried $M_n$ test Abel's test Dirichlet's test but geeting no where

Please help

Answer 1

It converges uniformly by Abel's test: the series $\sum_{n=1}^\infty\frac{(-1)^n}n$ converges uniformly and the sequence $\left(\frac{x^2+n^2}{n^2}\right)_{n\in\mathbb Z_+}$ is an uniformly bounded decreasing sequence of functions.

Answer 2

You can separate out $\sum (-1)^{n-1} \frac 1n$ (which is convergent) . The question reduces to uniform convergence of $\sum (-1)^{n-1} \frac {x^{2}} {n^{3}}$ and this series converges uniformly and absolutely by M-Test since $\sum \frac 1 {n^{3}} <\infty$.