I've been asked to show $\displaystyle\sum_{k=1}^{\infty}\frac{x^2}{k^2+x^2}$ $\forall x \in \mathbb{R}$ does not uniformly converge in $\mathbb{R}$. I do know and have already shown it plainly converges by the comparison test.
What I'm having with trouble is finding a value of $\epsilon$ that breaks the cauchy definition. We'd like to chose some $\epsilon$ so that $|\displaystyle\sum_{k=N+1}^{M}\frac{x^2}{k^2+x^2}|<\epsilon$ does not hold. Typically this involves chosing a value for $x$ smartly that forces the sum to just be some number, and then from there we can choose $\epsilon$ accordingly. These types of problems in general give me a hard time.
EDIT1: Having a hard time believing the proof given below, not sure if it holds for the chosen x.
EDIT2: Soo I think after meeting with some friends we were able to figure a much quicker way to do it than omnomnom's method below. A sequence of functions converges uniformly iff it is uniformly cauchy. Let
$S_{n}=\displaystyle\sum_{k=1}^{n}\frac{x^2}{k^{2}+x^{2}}$
and observe $|S_{n+1}-S_{n}|=\left|\displaystyle\frac{x^2}{(n+1)^2+x^2}\right|$. Now let $x=n+1$, so we have
$|S_{n+1}-S_{n}|=1/2$, Thus if we choose $\epsilon=1/4$ to begin with, we are done. I forgot that we could use the weaker version that takes advantage of consecutive terms of a sequence, and here it is quite helpful to do so. The appreciation I have in the below proof is mostly with the use of the Riemann sums. It's very neat and a useful thing to think about where applicable.
Using the Cauchy definition is one way to go about this. Another way is to define $$ f(x)=\sum_{k=1}^{\infty}\frac{x^2}{k^2+x^2} $$ And note that $$ |f(x)-f_N(x)|= \left|f(x)-\sum_{k=1}^{N}\frac{x^2}{k^2+x^2}\right|= \left|\sum_{k=N+1}^{\infty}\frac{x^2}{k^2+x^2}\right| $$ The key would then be to find a lower bound on this sum that depends on $x$ and $N$. In particular, we have $$ \begin{align} \left|\sum_{k=N+1}^{\infty}\frac{x^2}{k^2+x^2}\right| &= \sum_{k=N+1}^{\infty}\frac{x^2}{k^2+x^2}\\ &> \int_{N+1}^\infty \frac{x^2}{k^2+x^2}dk\\ &= x\int_{(N+1)/x}^\infty \frac{1}{1+u^2}du\\ &= x\left(\frac{\pi}{2}-\arctan\left(\frac{N+1}{x}\right)\right) \end{align} $$ We find this inequality by noting that the left Riemann sum overestimates the integral of decreasing functions.
From there, we can show that for a given $N$, the difference between the $N^{th}$ partial sum and $f(x)$ can be made arbitrarily large by the selection of a sufficiently large $x$.
In particular, take $\epsilon=1$. Consider any $N\in\mathbb N$. We note that for $x>1$, we have $$ \frac{\pi}{2}-\arctan\left(\frac{N+1}{x}\right) > \frac{\pi}{2}-\arctan\left(N+1\right) > 0 $$ Now, choose $x$ so that $x>1$ and $x>\left(\frac{\pi}{2}-\arctan\left(N+1\right)\right)^{-1}$. We then have $$ \begin{align} |f(x)-f_N(x)|&>x\left(\frac{\pi}{2}-\arctan\left(\frac{N+1}{x}\right)\right)\\ &> x\left(\frac{\pi}{2}-\arctan\left(N+1\right)\right)\\ &> \left(\frac{\pi}{2}-\arctan\left(N+1\right)\right)^{-1} \cdot \left(\frac{\pi}{2}-\arctan\left(N+1\right)\right)\\ &=1=\epsilon \end{align} $$ Thus, $f_n$ does not converge uniformly to $f$.