Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$

Question

I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.

However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.

Does non-uniform convergence of individual terms imply non-uniform convergence of series?

Answer 1

It's overkill in this case since the series $$ \sum_{n=1}^\infty \frac{nx}{1+n^2x^2} $$ is not even pointwise convergent for $x\neq 0$, but in general:

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If you have uniform convergence over $E$ of a series with partial sum $S_n=\sum_{k=1}^n f_n$ to a limit function $S$, then $$\begin{align*} \sup_{x\in E}|f_n(x)| &= \sup_{x\in E} |S_{n}(x)-S_{n-1}(x)| \leq \sup_{x\in E} |S_{n}(x)-S(x)|+ \sup_{x\in E} |S_{n-1}(x)-S(x)| \\&\xrightarrow[n\to\infty]{} 0+0=0 \end{align*}$$ so $(f_n)_n$ must converge uniformly (on $E$) to the zero function.

Answer 2

Say $\sum f_n(x) $ is uniformly convergent. Then there exists an $e$ such that for all x there exists integer N such that $$|f_{n+1}(x)+ f_n(x)|<e \forall n\ge N$$

Taking $e=1/3$ and $n=m>N$ and $x=1/m$ we get a contradiction. Hence series isn't uniformly convergent.

Answer 3

A simple comparison test shows the divergence of the series for $x\neq 0$:

For $x\neq 0$ you have $\frac 1{xn}\stackrel{n\to\infty}{\longrightarrow}0$. Hence, for $N$ large enough you can estimate

$$\sum_{n=N}^{\infty}\frac{nx}{1+n^2x}=\sum_{n=N}^{\infty}\frac{1}{\frac 1{nx}+n}\geq \sum_{n=N}^{\infty}\frac{1}{1+n}=\infty$$

Answer 4

Let $f_n(x)=\frac{n x}{1+n^2 x}$, the series $\sum_{n=1}^{\infty} \frac{n x}{1+n^2 x}$ is not defined on $x=\frac{-1}{n^2},\,\, n=1,2,3,...$. and for all other values of $x$, we have $\frac{n x}{1+n^2 x} \to 0,\,\, as\,\, n \to 0$.

According to M-test theorem, the series convergent uniformly on any bounded interval $(a,b],\,\, b >a>0$ for

$$\frac{n x}{1+n^2 x} \le \frac{b} {a n^2} $$

and it is clear that $$\frac{b}{a}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{b \pi^2}{6 a}.$$

The series diverges at $x= \frac{1}{1-n^2},\,\,\, n=1,2,3,...$ because the expression $\frac{n x}{1+n^2 x} =1$ at $x=\frac{1}{1-n^2}$ and the infinite summation for the last diverges.