1. ## Summon the Heroes-2

$\displaystyle Problem$ $\displaystyle 2$

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

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

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

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

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

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

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

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

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

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

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

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

4. Ingenious Work !!