Series Convergence of $ \sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}f(x)e^{-nx}dx$ 
Does the series
  $$ \sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}f(x)e^{-nx}dx$$
  with $f$ continuous on  $x\in [0,1]$
converge (absolutely)?

Do I have to calculate the integral, or can I use some other property of the integral?
 A: See that $f$ is bounded as continuous function on $[0,1]$ and 
$$\frac{1}{n}\int_{0}^{1}e^{-nx}dx = \frac{1}{n^2}(1-e^{-n})\le \frac{1}{n^2}$$
we have 
$$\sum_{n=1}^{\infty}\left|\frac{1}{n}\int_{0}^{1}f(x)e^{-nx}dx\right| \le M\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}e^{-nx}dx \le M\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$
A: Since $f$ is continuous on $[0,1]$ there is some constant $C$ such that $|f(x)|\leq C$ for all $x\in[0,1]$, hence
$$\Big|\int_0^1f(x)e^{-nx}\;dx\Big|\leq C\int_{0}^1e^{-nx}\;dx=C\frac{1-e^{-n}}{n}\leq \frac{C}{n}$$
Therefore the sum converges absolutely.
A: Your series is absolutely convergent.  There is no need for many calculations. Notice that the supremum norm of $f$ exists due to the continuity of $f$ on a compact set. You can see the absolutely convergence  by using the following:
\begin{align} 
\sum_{n\geq 1}\frac{1}{n}\bigg | \int^1_0 f(x) e^{-nx} dx\bigg |&\leq \sum_{n\geq 1} \frac{\Vert f\Vert_\infty}{n}\int^1_0e^{-nx} dx\\&=\sum_{n\geq 1}\frac{\Vert f\Vert_\infty}{n^2}(1-e^{-n}) 
\end{align} 
As usual $\Vert f\Vert_\infty :=\sup_{x\in [0,1]}|f(x)|$
