1. ## Summon the Heroes-2

$Problem$ $2$

Assume $\sum_{n=1}^{\infty}a_{n}$ converges

where $a_{n}>0$ and $a_{n} \geq a_{n+1}$ for $\forall$ $n \in \mathbb{N}^+$

show that: $\lim_{n \rightarrow \infty}na_{n}=0$

2. Originally Posted by Xingyuan
Assume $\sum_{n=1}^{\infty}a_{n}$ converges

where $a_{n}>0$ and $a_{n} \geq a_{n+1}$ for $\forall$ $n \in \mathbb{N}^+$

show that: $\lim_{n \rightarrow \infty}na_{n}=0$
Suppose that $\sum_{n=1}^{\infty}a_{n} = S$. Given $\varepsilon>0$, choose N so that $\sum_{r=1}^{N}a_r > S-\tfrac12\varepsilon$. Notice that if $n>2N$ it follows that $n-N>\tfrac12n$.

If $n>2N$ then $\tfrac12\varepsilon > \sum_{r=N+1}^{n}a_r >(n-N)a_n$ (since the terms $a_r$ are decreasing), and so $\tfrac12\varepsilon > \tfrac12na_n$. Thus $n>2N\;\Longrightarrow\;na_n<\varepsilon$. Therefore $na_n\to0$ as $n\to\infty.$

3. I've found a nice proof without $\varepsilon$'s...

Note that the sequence of general term $u_n=\sum_{k=1}^n (a_k-a_n)$, $n\geq 1$, is nondecreasing. This is because the terms of the sum are nonnegative ( $a_n\leq a_k$ for $k\leq n$), and each of them (i.e. $a_k-a_n$) increases with $n$.

Plus, $u_n\leq \sum_{k=1}^\infty a_k<\infty$, hence $(u_n)_{n\geq 1}$ converges to a finite limit.

However, $u_n=\sum_{k=1}^n a_k - na_n$ and the sum converges to a finite limit as $n\to\infty$, hence $na_n$ must converge to a finite limit $\alpha$ as well.

This limit has to be 0, for otherwise $a_k\sim \frac{\alpha}{k}$ is the general term of a divergent series. Thus, $n a_n\to 0$.

4. Ingenious Work !!