TITLE: Why does the tail $a_N+a_{N+1}+a_{N+2}\ldots$ of convergent series $\sum a_n$ tend to $0$ as $N\to\infty$?
QUESTION [7 upvotes]: Let $a(n)>0$ for all $n \in \mathbb{N}$ be such that $\sum a(n)$ converges. Define $r(n):=\sum\limits_{k=n+1}^\infty a(k)$. The claim is $$\lim_{n \to \infty} r(n) = 0,$$ but I cannot see how to prove it. Please help.

REPLY [6 votes]: We know that the series $$\sum_{k=1}^\infty a(k)$$ converges, meaning by definition that its sequence of partial sums $$s(n):=\sum_{k=1}^na(k)$$ converges. Take $\varepsilon>0$, so by convergence of its sequence of partial sums, we have for some $N$ and all $n\geq N$ that $$\left|s(n)-\sum_{k=1}^\infty a(k)\right|<\varepsilon.$$ But $$s(n)-\sum_{k=1}^\infty a(k)=-r(n),$$ so $-r(n)\to 0$, and so $r(n)\to 0$.

REPLY [3 votes]: Let $s(n) = \sum _{k = 0}^n a(k)$.
Let $s = \sum _{k = 0}^\infty a(k)$.
Then $\sum _{k = n+1}^\infty a(k) = s - s(n)$.
Hence $\lim_{n\rightarrow\infty} \sum _{k = n+1}^\infty a(k) = \lim_{n\rightarrow\infty} s - s(n) = 0$