Determining whether this infinite sum goes to zero I have a sum which I know is finite of the form $\sum_{-\infty}^{\infty} f(x) <\infty$.
I wish to determine whether the following holds:
$\lim_{n \to \infty} \frac{1}{n} \sum_{-n}^n |x|f(x)=0$.
My instinct is that it does since for all $x$, we know that $\frac{x}{n}<1$ and since the sum is finite, each summand should go to zero and $n \to \infty$.
However I'm not sure how to prove this since I have no knowledge of what $f(x)$ is other than the property $f(x)=f(-x)$
 A: Suppose you have a sequence $(f(k):k\in\mathbb{Z})$ with $f(k)\geq 0$ for every $k$ and $\displaystyle{\sum_{-\infty}^{\infty}f(k):=\lim_{n\to\infty}\sum_{k=-n}^n f(k)=L<\infty}$. Then, for every $n$ and fixing $m\geq 2$ you have:

Claim: $\displaystyle{\lim_{n\to\infty}\sum_{|k|>n/m}f(k)=0}$.

Proof of the Claim: Notice that since $\sum_{k=-\infty}^\infty f(k)$ is a convergent series, we must have that the sequence $S_n=\sum_{k=-n}^n f(k)$ is a convergent sequence, with limit $L=\sum_{k=-\infty}^\infty f(k)$. 
Thus, given $\epsilon>0$ there is $N$ such that whenever $n\geq N$, $$|L-S_n|=\sum_{k=-\infty}^\infty f(k) - \sum_{k=-n}^n f(k)=\sum_{|k|>n}<\epsilon.$$
Now, if we choose $N_1=Nm$, then whenever $n\geq N_1=Nm$, $n/m\geq N$ and we have $$\left|\sum_{k=-n/m}^{n/m}f(k)\right|<\epsilon$$ which shows that $\lim_{n\to\infty}  \sum_{|k|>n/m}f(k)=0$.   $\hspace{1cm}\Box$
Now, for every fixed $m\in\mathbb{N}$, we have:
\begin{align*}0\leq \dfrac{1}{n}\sum_{k=-n}^n |k|\cdot f(k)&=\sum_{k=-n}^n\dfrac{|k|}{n}f(k)=\sum_{k=-n/m}^{n/m} \dfrac{|k|}{n}f(k) + \sum_{|k|>n/m}\dfrac{|k|}{n}f(k)\\
&\leq \sum_{k=-n/m}^{n/m} \dfrac{|n/m|}{n}f(k) + \sum_{|k|>n/m}\dfrac{|n|}{n}f(k)\\
&=\dfrac{1}{m}\sum_{k=-n/m}^{n/m}f(k)+\sum_{|k|>n/m}f(k).
\end{align*}
and taking the limit we obtain
\begin{align*}
\lim_{n\to\infty}0&\leq \lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-n}^n |k|f(k)\leq \dfrac{1}{m}\lim_{n\to\infty} \left(\sum_{k=-n/m}^{n/m} f(k)\right) + \lim_{n\to\infty}\left(\sum_{|k|>n/m}f(k)\right)\\
0&\leq\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-n}^n |k|f(k)\leq \dfrac{L}{m}+0.
\end{align*}
and since this equality is true for all $m\in\mathbb{N}$ and $L<\infty$, we conclude that $\displaystyle{\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-n}^n |k|f(k)=0}$.

Edit: If you do not assume $f(k)\geq 0$, notice that the Claim is still valid, and you can finish by doing 
\begin{align*}
0&\leq \lim_{n\to\infty} \left|\dfrac{1}{n}\sum_{k=-n}^n |k|f(k)\right|\\
&\leq \lim_{n\to\infty} \dfrac{1}{m}\left|\sum_{k=-n/m}^n f(k)\right| + \lim_{n\to\infty} \left|\sum_{k=-n}^n |k|f(k)\right|
&\leq \dfrac{|L|}{m}
\end{align*}
Since this is true for every $m\in\mathbb{N}$, we end up concluding that $$\lim_{n\to\infty} \left|\dfrac{1}{n}\sum_{k=-n}^n |k|f(k)\right|=0,$$ as desired.
