Showing a series converges non-uniformly For the series $$f(x) = \sum_{n=0}^\infty \frac{1}{n(1+nx^2)}$$ my lecture notes use the Weierstrass M-test to show that this converges uniformly on any interval of the form $(-\infty,-a)$ or $(a,\infty)$ for $a>0$ and says that it converges non uniformly on $(0,a)$ or $(-a,0)$ for any $a>0$. 
Isn't this a contradiction? I.e. can't you find two such $a$'s where it converges uniformly and non-uniformly. 
Also how does one show that there is some interval where this converges non-uniformly? 
Thanks!
 A: Define
$$
f_k(x)=\sum_{n=0}^{k-1}\frac{1}{n(1+nx^2)}\tag{1}
$$
For $|x|>a$, we have
$$
\begin{align}
|f(x)-f_k(x)|
&=\sum_{n=k}^\infty\frac{1}{n(1+nx^2)}\\
&<\sum_{n=k}^\infty\frac{1}{n(1+na^2)}\\
&<\frac{1}{a^2}\sum_{n=k}^\infty\frac{1}{n^2}\\
&<\frac{1}{a^2}\sum_{n=k}^\infty\left(\frac{1}{n-1/2}-\frac{1}{n+1/2}\right)\\
&=\frac{1}{a^2(k-1/2)}\tag{2}
\end{align}
$$
Estimate $(2)$ says that $f_k\to f$ uniformly on $\{x:|x|>a\}$.
However,
$$
\begin{align}
|f(x)-f_k(x)|
&=\sum_{n=k}^\infty\frac{1}{n(1+nx^2)}\\
&\ge\sum_{n=k}^{1/x^2}\frac{1}{n(1+nx^2)}\\
&\ge\frac12\sum_{n=k}^{1/x^2}\frac{1}{n}\\
&\ge-\frac12\log(x^2(k+1))\tag{3}
\end{align}
$$
That is, for any positive integer $k$, if $0<x<\frac{1}{e\sqrt{k+1}}$, then $|f(x)-f_k(x)|\ge1$. Thus, the convergence is not uniform on any neighborhood of $x=0$.
A: First, you need to start your sum at $n=1$, not $n=0$.
To address your first question:
There is no contradiction. If a series converges uniformly on a set $O$, then it converges uniformly on any subset of $O$; but, the series does not necessarily converge uniformly on a set $U\supset O$.
You have uniform convergence on a set $(a,\infty)$, $a>0$ and non-uniform convergence on a larger set $(0,\infty)$, which is ok...

To, hopefully, address your second question:
In your example, you have uniform convergence on sets of the form $(a,\infty)$, $a>0$. 
You have pointwise convergence on the set $(0,\infty)$.  Moreover, since the series diverges at $x=0$, this is the largest interval, unbounded on the right, for which you have pointwise convergence.  This is then 
the natural candidate for an interval over which you could have non-uniform convergence.
Moreover, things would have to go awry near $x=0$.
But is this indeed the case?
Let's look at the graphs of the first few partial sums $s_n=\sum\limits_{k=1}^n {1\over k(1+kx^2). }$

We are led to suspect that the series does not converge uniformly on $(0,\infty)$. (look at how the difference between the $s_k$ grows larger as $x\rightarrow0^+$).
In fact, given $N$, we may choose $M>N$ so that $\sum\limits_{n=N}^M{1\over n}\ge 1$. Then
$$
\lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n(1+nx^2)}
=\lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n}\ge1.
$$ 
So the series $\sum\limits_{n=1}^\infty {1\over n(1+nx^2)}$ is not uniformly Cauchy on $(0,\infty)$ and thus not uniformly convergent on $(0,\infty)$. 
Since $x=0$ is the "bad point" as far as uniform convergence is concerned, the series does not converge uniformly on any interval of the form $(0,a)$.
A: There is no contradiction. Note that $a>0$ is arbitrary. This series converges uniformly on any interval of the form $(-\infty,-a)$ or $(-\infty,-b]$ or $(a,\infty)$ or $[b,\infty)$ where $a>0,b>0$. Due to 
$$|\frac{1}{n(1+nx^2)}|\leq \frac{1}{n(1+na^2)}$$
(the series $\sum\frac{1}{n(1+na^2)}$ converges), so by Weierstrass M-test, the form series converges uniformly. For the other hand, give you any interval $0\in(a,b)$, take $x=0$ , then the series becomes $\sum\frac{1}{n}$, obviously, this series does not converge.
