Let me suggest the question in my lecture's note.
Show this series uniformly convergent on $\mathbb{R}$
$\sum_{n=0} ^{\infty} {1 \over 1+(n+x)^2}$
One thing sure, we can easily show it by using Weierstrass thm that by $\sum {1 \over 1+(n+x)^2} \leq \sum M_n$ (Here the $M_n$ only depend on the variable $n$) Here is my trial for finding the $M_n$
I found my lecture's solution in his note. He said
Say $a>0$, $\vert n \vert > 4a$ for $a \in \mathbb{Z}$.
Hence, $\forall x \in [-a,a]$, ${1 \over 1+(n+x)^2} < {1 \over 1+ n^2/2}$
First, I can't guess how could he discover the $n$ s.t. $\vert n \vert > 4a$. So, I tried different method like the below
${1 \over 1+(n+x)^2} < {1 \over (n+x)^2} < {1 \over 4nx}$
For taking the $M_n$, only left step is getting rid of the $x$. What should I do next step?
Any help would be appreciated.
A necessary condition for uniform convergence of a series $\sum f_n(x)$ for $x \in D$ is that $f_n(x) \to 0$ uniformly as $n \to \infty$. This follows from the Cauchy criterion, and is equivalent to $\lim_{n \to \infty} \sup_{x \in D} |f_n(x)| = 0.$
In this case, we have with $x_n = -n+1/n \in D=(-\infty,0),$
$$\sup_{x \in D }|f_n(x)| =\sup_{x \in D} \frac{1}{1+(n+x)^2} \geqslant \frac{1}{1+(n+ x_n)^2} = \frac{n^2}{n^2+1} \to 1$$
Since the RHS does not converge to $0$ the convergence cannot be uniform on $(-\infty,0)$.
As shown in a comment convergence on $[0,\infty)$ is, in fact, uniform by the Weierstrass M-test.