Swapping integral and sum using dominated convergence theorem 
Show that
  $$
\sum_{n=1}^\infty \int_0^\infty t^{s/2-1}e^{-\pi n^2t}dt
 = \int_0^\infty t^{s/2-1} \sum_{n=1}^\infty e^{-\pi n^2t} dt
$$
  with $s > 1$ using the dominated convergence theorem.

If I have understood correctly, I can define $f_k:= \sum_{n=1}^k t^{s/2-1}e^{-\pi n^2t}$ which I have already been able to show integrable and that it converges pointwise ($f = \lim_{k \rightarrow \infty} f_k$). Now the last part of the requirements of the theorem: I struggle to find an integrable function $g$ with $|f_k| \leq g$ for all $k \in \mathbb{N}$.
 A: As mentioned in the comments, what you are trying to do is overkill since monotone convergence is enough. But if you want to use dominated convergence you can do
\begin{align}
|f_k|&= \sum_{n=1}^k t^{s/2-1}e^{-\pi n^2t}=t^{s/2-1}e^{-\pi t}\,\sum_{n=1}^k e^{-\pi (n^2-1)t}\\ \ \\
&\leq t^{s/2-1}e^{-\pi t}\,\sum_{n=1}^\infty e^{-\pi (n^2-1)t}\\ \ \\
&\leq t^{s/2-1}e^{-\pi t}\,\sum_{n=0}^\infty e^{-\pi nt}\\ \ \\
&=t^{s/2-1}e^{-\pi t}\,\frac1 {1-e^{-\pi t}}\\ \ \\
\end{align}
A: Following up on @zugzug's comment, we can prove this using the monotone convergence theorem. Let $f_n(t) = \sum_{k=1}^n t^{s/2-1}e^{-k^2\pi t}$ for $n=1,2,\ldots$. Then $f_n(t)\geqslant 0$ and
\begin{align}
f_{n+t}(t) - f_n(t) &=  \sum_{k=1}^{n+1} t^{s/2-1}e^{-k^2\pi t} - \sum_{k=1}^n t^{s/2-1}e^{-k^2\pi t}\\
&= t^{s/2-1}e^{-k^2\pi t} >0
\end{align}
for all $t\geqslant 0$, so the limit $f:=\lim_{n\to\infty} f_n$ is measurable and
\begin{align}
\int_0^\infty \lim_{n\to\infty} f_n(t)\ \mathsf dt&=
 \int_0^\infty t^{s/2-1}\sum_{n=1}^\infty e^{-n^2\pi t}\ \mathsf dt\\
 &= \lim_{n\to\infty} \int_0^\infty f_n(t)\ \mathsf dt\\ &= \sum_{n=1}^\infty \int_0^\infty  f_n(t)\ \mathsf dt\\ &= \sum_{n=1}^\infty \int_0^\infty t^{s/2-1} e^{-n^2\pi t}\ \mathsf dt.
\end{align}
