If $a_n \geq 0$ $\forall n \in \mathbb{N}$ and $\sum_{n=1}^ \infty a_n < \infty$, what can be said about $\lim_{k\to\infty}\sum_{n=k}^\infty a_n$?
My claim is that $\lim_{k\to\infty}\sum_{n=k}^\infty a_n = 0$. To see this, first note that for all $k\in \mathbb{N}$, the series $\sum_{n=k}^\infty a_n$ converges. Write $L_k = \sum_{n=k}^\infty a_n$ for each $k \in \mathbb{N}$. Then we have $$ L_{k+1} = \sum_{n=k+1}^\infty a_n \leq a_k + \sum_{n=k+1}^\infty a_n = \sum_{n=k}^\infty a_n = L_k. $$ So the sequence $\{L_k\}$ is monotone decreasing. Moreover, $0 \leq L_k$ for all $k\in \mathbb{N}$. From these two facts, we can apply the monotone convergence theorem for sequences which tells us that
$$ \lim_{k \to \infty} L_k \to \inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\}. $$
So it suffices to show that $$ \inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\} =0. $$ Let $\epsilon >0$. Since $\sum_{n=1}^\infty a_n$ converges, we can choose $N \in \mathbb{N}$ such that for all $j > i \geq N$ we have $$ \sum_{n=i}^j a_n = \left| \sum_{n=i}^j a_n \right| < \epsilon. $$
Hence, for this $N$ we have
$$ \sum_{n=N}^\infty a_n = \lim_{j \to \infty}\sum_{n=N}^j a_n < 0 + \epsilon. $$ This shows that $$ \inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\} =0. $$
Can anyone critique my proof?
