Is $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$ continuous?. Where is bluff? 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 $A_{n}$ &lt; M. Thus, I can write $\lim_{x \rightarrow 0} f(x) &lt; lim_{x \rightarrow 0} xM = 0$
well. that would be finished, but as I have mentioned before, I suspect bluff.
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 $x&gt;0$ 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
$$
\int_0^{+\infty}e^{-xt^2}dt=\frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}}, \quad x&gt;0.\tag6
$$ 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 $(*)$.
Note the summation when $x=0$ is infinite since it is an infinite sum of 1's, thus contradicting your argument of finite $M$.