Is $\sum _{n=1} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 }$ uniformly convergent on $(-\pi , \pi)$?

Question

Is this series uniformly convergent on $(-\pi , \pi)$: $$\sum _{n=1} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 }\,?$$

My Attempt:

If the series were convergent we would have got a natural number $k$ for a fixed $\epsilon>0$ such that $\sum _{n=k} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 } < \epsilon$ for all $x \in (-\pi , \pi)$. But for this case we will get the term $f_n$ in the summation greater than $1$ for $-\pi + 1/n$. So for every $n$ we will get a $x\in (-\pi , \pi) $ such that summation at $x$ is greater than $1$.

That's why the series is not uniformly convergent on $(-\pi , \pi)$.

Can somebody please tell me if I have gone wrong anywhere?

Answer 1

Your answer is correct. The series isn't uniformly convergent.

Answer 2

If a series uniformly converges, then its summands must approach zero uniformly.

Clearly, the summands do not converge uniformly to $0$.

To see this, take $\epsilon=1$.

Then, for all $N$, take any $n>N$ and $x=-\pi +\frac1n$ and find that

$$\left|\frac{1}{\left(x+\pi\right)^2 n^2}\right|=\left|\frac{1}{\left(-\pi+\frac1n+\pi\right)^2 n^2}\right|=1$$

This negates the uniform convergence of the summand $\frac{1}{(x+\pi)^2n^2}$. And we are done!

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If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.

With $\epsilon=1$, we have for all $N>1$

$$\begin{align} \left|\sum_{n=1}^\infty \frac{1}{(x+\pi)^2n^2}-\sum_{n=1}^N \frac{1}{(x+\pi)^2n^2} \right|&=\sum_{n=N+1}^\infty \frac{1}{(x+\pi)^2n^2}\\\\ &\ge \frac1{(x+\pi)^2(N+1)^2}\\\\ &\ge1 \end{align}$$

whenever $x= -\pi+\frac1{ N+1}$. And we are done!

Answer 3

Put $f_n(x)=\displaystyle\sum_{k=1}^n\frac{1}{(x+\pi)^2k^2}$. Then for each $x\in(\pi,\pi)$, $f_n(x)\rightarrow \frac{\pi^2}{6(x+\pi)^2}$. Hence $f_n\xrightarrow{p.w} f$, where $f(x)=\frac{\pi^2}{6(x+\pi)^2}$ for all $x\in(-\pi,\pi)$.

For each $n\in\mathbb{N}$, let \begin{eqnarray} M_n&=&\sup\{|f_n(x)-f(x)|~|~x\in(\pi,\pi)\}\\ &=& \sup\left\{\left|\displaystyle\sum_{k=n+1}^{\infty}\frac{1}{(x+\pi)^2k^2}\right|~|~~x\in(\pi,\pi)\right\}\\ &=&\left(\displaystyle\sum_{k=n+1}^{\infty}\frac{1}{k^2}\right)\sup\left\{\left|\displaystyle\frac{1}{(x+\pi)^2}\right|~|~~x\in(\pi,\pi)\right\}\\ &=& \infty \end{eqnarray} i.e., $M_n=\infty$ for all $n$ and so $M_n$ does not converges to zero as $n$ tends to $\infty$. Hence $f_n$ does not converges uniformly.