Proving uniform convergence of this series

Question

I'm trying to prove that the following series

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

converges uniformly on every finite interval $I$ in $\mathbb{R}$.

In the previous exercice I've shown that the series is not absolute convergent for any $x \in \mathbb{R}$, so this might be of use, but I'm not sure how.

I've attempted to use the Weierstrass M-test and tried to find some $M_n \geq \dfrac{x^2 + n}{n^2} \; \forall x \in I$ such that $\sum M_n$ converges. But I realised that this would imply that $M_n \geq \frac{1}{n}$ and thus $\sum M_n$ cannot converge.

I also tried finding the partial sums for the series, but this also turned out to be pretty tough.

How would I go about showing uniform convergence on $I$?

Answer 1

First given $x \in \mathbb{R}$:

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

So if you prove $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent you are ready (since the second sum converges and is constant and not depends on $x$).

$x^2$ is bounded for every finite interval on $I$ by $M_I$:

$$\left|(-1)^n\frac{x^2}{n^2}\right|\leq \frac{M_I}{n^2} $$

So $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent.

Uniformity: pointwise $f_m(x)$ converges to:

$$f(x)=x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}$$ So: $$|f(x)-f_m(x)|=$$ $$\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$ $$\leq\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$

$$\leq|x^2|\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$ $$\leq M_I\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$

At that point you can complete the demostration (because the convergence is not given by the point $x$ you take)

Answer 2

Let $M>0$ such that $I \subset [-M,M]$

Let $$\begin{array}{ccccc} f & : & I & \to & \mathbb R \\ & & x & \mapsto & \sum_{n=1}^{\infty}(-1)^n \dfrac{x^2 + n}{n^2} \\ \end{array}$$

For $m\in \mathbb N$, let $$\begin{array}{ccccc} f_m & : & I & \to & \mathbb R \\ & & x & \mapsto & \sum_{n=1}^{m}(-1)^n \dfrac{x^2 + n}{n^2} \\ \end{array}$$

Now for $x \in I$, because the series is alternate, $$|f(x)-f_m(x)|\leq \dfrac{x^2 + n+1}{(n+1)^2} \leq \dfrac{M^2 + n+1}{(n+1)^2}$$

Because $\frac{M^2 + n+1}{(n+1)^2} \to 0$ and is independent on $x$, uniform convergence is granted.