When the starting index of a series goes to infinity, what happens? Let's say you have a series that looks like $\sum^\infty_{n=N}f(n)$, where $f(n)$ is some $n$-dependent thing. If you take the limit of this series as $N$ approaches infinity, what kind of stuff can you use to figure out what the limit is? For instance, I read somewhere else $\sum^\infty_{n=N}\frac{1}{n^2} \rightarrow 0$ as $N \rightarrow \infty$, but why?
 A: Because of the Cauchy-criterion of the convergent sequence. For the existence of  $\displaystyle\sum_{n=1}^{\infty}a_{n}$, it simply means that the sequence $s_{n}=\displaystyle\sum_{k=1}^{n}a_{k}$ is convergent, then it is Cauchy. Then for each $\epsilon>0$, we can find an $N$ such that $n,m\geq N$ implies that $|s_{n}-s_{m}|<\epsilon$. Suppose without loss of generality that $m>n$, so $|s_{n}-s_{m}|=\left|\displaystyle\sum_{k=n+1}^{m}a_{k}\right|<\epsilon$. Now we take $m\rightarrow\infty$ to get $\left|\displaystyle\sum_{k=n+1}^{\infty}a_{k}\right|\leq\epsilon$ for all such $n\geq N$, this means that $\displaystyle\sum_{k=n}^{\infty}a_{k}\rightarrow 0$ as $n\rightarrow\infty$.
A: Hopefully not too trivial:
Assume $\lim_{N \rightarrow \infty} \sum_{k=N}^{\infty}f(k)=0$:
$\epsilon/2$ given.
There is a $N_0$ s.t. for $N>N_0$
$|\sum_{k=N}^{\infty}f(k)|<\epsilon/2$.
For $m \ge n >N_0$
$|\sum_{k=n}^{m}f(k)|=$
$|\sum_{k=n}^{\infty}f(k)-\sum_{k=m}^{\infty}f(k)|<$
$|\sum_{k=n}^{\infty}f(k)|+|\sum_{k=m}^{\infty}f(k)| <$
$\epsilon/2+\epsilon/2=\epsilon$, i .e.
$S_n:=\sum_{k=1}^{n}f(k)$ is Cauchy, hence convergent.
