Prove a series is not uniform convergent ? $\sum_{n=1}^\infty n^2xe^{-n^2x}$

convergence-divergencereal-analysisuniform-convergence

We define the function $f_n(x), \, n=1,2,…$ by
\begin{align*}
f_n(x) = n^2xe^{-n^2x} \qquad \qquad(x\in \mathbb{R})
\end{align*}

  1. Show that the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent on $[a,\infty)$ if $a$ is a positive number.

  2. Show that the series $\sum_{n=1}^\infty f_n(x)$ is not uniformly convergent on $[0,\infty)$

Idea: The convergence of series is defined in terms of the convergence of its sequence of partial sums. Here we use the Weierstrass M-test to check the uniform convergence.
Weierstrass M-test : Let $\{f_n\}$ be a sequence of function $f_n: A\xrightarrow[]{} \mathbb{R}$ and suppose that $\forall n \in \mathbb{Z}^+$ there exists a constant sequence $M_n$ such that,
\begin{align*}
|f_n(x)| \leq M_n \qquad\qquad \forall x\in A
\end{align*}
If the series $\sum_{n=1}^\infty M_n$ is convergent then $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Solution:

  1. First find the $M_n$ using the Taylor series expansion for $e^{n^2x}$
    \begin{align*}
    e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + …\\
    e^{n^2x} &= 1 + n^2x + \frac{(n^2x)^2}{2!} + \frac{(n^2x)^3}{3!} + …\\
    e^{n^2x} &> \frac{(n^2x)^2}{2!}\\
    \frac{1}{e^{n^2x}} &< \frac{2!}{(n^2x)^2}\\
    \bigg|\frac{n^2x}{e^{n^2x}}\bigg| &< \bigg|\frac{2!}{n^2x}\bigg| \leq \frac{2}{n^2a} \qquad\qquad \text{where} \,\,\, a > 0
    \end{align*}

Take $M_n = \frac{2}{n^2a}$ and we need to check $\sum_{n=1}^\infty M_n$ is convergent on the given intervals
\begin{align*}
\sum_{n=1}^\infty M_n =\sum_{n=1}^\infty \frac{2}{n^2a} = \frac{2}{a} \sum_{n=1}^\infty \frac{1}{n^2}
\end{align*}
We know by the p-series $\sum_{n=1}^\infty \frac{1}{n^p}$ where $p>1$ then the series is convergent. So in this case $p=2$ therefore the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent. Then the series $\sum_{n=1}^\infty M_n$ is convergent on $[a,\infty)$ therefore by the Weierstrass M-test the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Is my answer for (1) correct? I tried to find the answer for (2) but failed. I could not find the partial sum of this series. Is there any other ways to prove that, can anyone give a little explanation I am really stuck here. Thanks

Best Answer

On $\mathbb{R}^+$, $g(x)=x e^{-x}$ is a positive function attaining its absolute maximum, $\frac{1}{e}$, at $x=1$.
It follows that $f_n(x)=g(n^2 x)$ attains its absolute maximum, $\frac{1}{e}$, at $x=\frac{1}{n^2}$.

If $[a,+\infty)$ we have that $\sum_{n\geq 1}f_n(x)$, except for the first $\left\lceil \frac{1}{\sqrt{a}}\right\rceil$ terms, is a series of continuous and decreasing functions, with both $\lim_{x\to a}$ and $\lim_{x\to +\infty}\sum_{n\geq 1}f_n(x)$ being finite.

On the other hand we cannot have uniform convergence on $[0,+\infty)$ since $f_n(0)=0$ but the value of $\sum_{n\geq 1}f_n(x)$ is greater than $\frac{1}{e}$ for any $x$ of the form $\frac{1}{n^2}$.

As shown here through Gauss circle method we have $$ \sum_{n\geq 1}e^{-n^2 x}\sim -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}} $$ as $x\to 0^+$, hence $$ \sum_{n\geq 1} n^2 x e^{-n^2 x}\sim \frac{\sqrt{\pi}}{4\sqrt{x}} $$ as $x\to 0^+$. The series is unbounded in a right neightbourhood of the origin.

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