Is $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$ continuous?. Where is bluff?

Question

I have a function defined by $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$. The task is to check, whether $f(x)$ is continuous at $x = 0$. I have proposition of a solution and I would like someone to point out a bluff as there most likely is one - solution is suspisciously too short. \begin{equation} f(x) = \sqrt(x) \sum^{\infty}_{n=1} e^{-n^2x} = \sqrt(x) A_{n} \end{equation} I do it because x doesn't change when summing, I treat it as constant. The sum $A_{n}$ is obviously convergent, so there exist some finite M conforming $A_{n}$ < M. Thus, I can write $\lim_{x \rightarrow 0} f(x) < lim_{x \rightarrow 0} xM = 0$ well. that would be finished, but as I have mentioned before, I suspect bluff.

Answer 1

Here is an elementary approach to obtain the desired limit.

As $x$ tends to $0^+$, we have

$$\sqrt{x}\sum_{n=1}^{+\infty}e^{\large -xn^2} \longrightarrow \frac{\sqrt{\pi}}{2}. \tag{*}$$

Proof. Let $x>0$ and $n\geq 0$. Since $\displaystyle t \in [0,+\infty) \rightarrow e^{\large -xt^2}$ is a decreasing function, we have $$ e^{-x(n+1)^2} \leq e^{-xt^2} \leq e^{-xn^2}, \quad t \in [n,n+1], \tag1 $$ integrating $(1)$ with respect to $t$, we get $$ \int_n^{n+1}e^{-xt^2}dt \leq \int_n^{n+1}e^{\large -x n^2}dt = e^{\large -x n^2}\tag2 $$ and $$ \int_{n-1}^{n}e^{\large -x n^2}dt = e^{\large -x n^2}\leq \int_{n-1}^{n}e^{-xt^2} dt. \tag3 $$ Then, summing $(2)$ for $n\geq0$, gives $$ \int_0^{+\infty}e^{-xt^2}dt \leq 1+\sum_{n=1}^{+\infty}e^{\large -x n^2} \tag4 $$ and summing $(3)$ for $n\geq 1$, gives $$ \sum_{n=1}^{+\infty}e^{\large -x n^2} \leq \int_0^{+\infty}e^{-xt^2}dt. \tag5 $$Recalling the gaussian integral evaluation, we have $$ \int_0^{+\infty}e^{-xt^2}dt=\frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}}, \quad x>0.\tag6 $$ Hence, combining $(4)$, $(5)$ and $(6)$, gives $$ \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}}-1 \leq \sum_{n=1}^{+\infty}e^{\large -x n^2} \leq \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}} \tag7 $$ leading, as $x$ tends to $0^+$, to the announced result $(*)$.

Answer 2

Note the summation when $x=0$ is infinite since it is an infinite sum of 1's, thus contradicting your argument of finite $M$.