TITLE: Convergence of a positive series, given that $\sum_{k=1}^n(a_k-a_n)$ is bounded
QUESTION [5 upvotes]: Question:
Assume $\sum_{n=1}^\infty a_n$ is a positive series, with $a_n$ decreases to $0$, and
$$\sum_{k=1}^n(a_k-a_n)$$ is bounded with respect to $n$. Prove that $\sum_{n=1}^\infty a_n$ is convergent.

My idea:
Suppose
$$(a_1-a_n)+(a_2-a_n)+\cdots+(a_n-a_n)\leq M$$ for any integer $n$.
Then for any integer $m$, choose a greater $n$ to have
$$\begin{align}&\quad\,(a_1-a_n)+(a_2-a_n)+\cdots+(a_m-a_n)\\
&\leq (a_1-a_n)+(a_2-a_n)+\cdots+(a_m-a_n)+\cdots+(a_n-a_n)\\
&\leq M\end{align}$$ since every term in brackets is nonnegative.
Let $n\to\infty$ in $$(a_1-a_n)+(a_2-a_n)+\cdots+(a_m-a_n)\leq M$$ to conclude $a_1+a_2+\cdots+a_m\leq M$. Then let $m\to\infty$ to conclude that $\sum a_n$ is convergent.
Is that a reasonable proof? Thank you!!

REPLY [3 votes]: Since $(a_n)$ is decreasing, note that $\forall n\geq 1$, $ \sum_{k=1}^n(a_k-a_n)\geq 0$.
Let $S_n = \sum_{k=1}^n a_k$ and note that for $n\geq 2$, $$\sum_{k=1}^n(a_k-a_n) = S_n - n(S_n-S_{n-1}) = n(n-1) \Big( \frac{S_{n-1}}{n-1}-\frac{S_{n}}{n} \Big). $$
By the boundedness assumption and the non-negativity, there is some $M\geq 0$ such that for all $k\geq 2$,
$$0\leq \frac{S_{k-1}}{k-1}-\frac{S_{k}}{k} \leq \frac M{k(k-1)}$$
Let $n\leq m$ be integers $\geq 2$ and sum the previous inequality from $k=n$ to $k=m$ to get
$$0\leq \frac{S_{n-1}}{n-1}-\frac{S_{m}}{m}\leq M(\frac{1}{n-1}-\frac{1}{m}). $$
Since $a_k\to 0$, Cesàro's theorem gives $\lim_{m\to \infty} \frac{S_{m}}{m} = 0$. Thus, letting $m\to \infty$ yields $$0\leq \frac{S_{n-1}}{n-1} \leq \frac{M}{n-1},$$
hence $\sum_{k=1}^n a_k = S_n$ is bounded by $M$. Since the $a_n$ are non-negative, the series is convergent.