Prove $\lim\limits_{n \to \infty} na_n=0.$ 
Suppose $\sum_{n=1}^{\infty} a_n$ be a convergent positive series, and
satisfy $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_n}=1$. Prove
$\lim\limits_{n \to \infty} na_n=0.$

First, we may consider applying Cauchy's condensation test. Since $\sum_{n=1}^{\infty}a_n$ is convergent, $\sum_{n=1}^{\infty} 2^n a_{2^n}$ is convergent as well, which implies $\lim\limits_{n \to \infty}2^n a_{2^n}=0$. Hence, there already exists a subsequence of $\{na_n\}$ convergent to zero. But how to go on ?
 A: This result is wrong.

Counter Example
For all $k \ge 0; 0 \le m \le 2^{k}-1$, define $a_{2^k+m}$ as follows:
$$ a_{2^k+m}= 2^{-k}.\max \left( 2^{-m\frac{k^2}{2^k} } ,2^{-1-(2^k-m)\frac{k^2}{2^k} }\right) $$
This defintion is nothing extraordinary if you look at the line graph of the log sequence $\left( \log(a_n), n \ge 1\right)$ . It is just a piecewise linear curves whose maximas are attained at points of form $2^k$.
Also, the slope of that graph becomes more flatter and eventually becomes a horizontal lines.( Because $\frac{k^2}{2^k} \rightarrow 0$ ).
And we can even check it by some simple algebraic arguments that:
$$ \lim \frac{a_n}{a_{n+1}}= 1$$
While we can check that :
$$ \sum_{ m \ge 0}^{2^k-1} a_{2^k+m} \le \left(2^{-k}+2^{-k-1} \right)\left( 
 \frac{ 1-2^{-k^2}}{1-2^{-k^2/2^k}} \right)$$
( replacing the max by the sum of each components)
$$ LHS \le 2.2^{-k}. \underbrace{ \frac{4.2^k}{ k^2 } }_{1-2^{-k^2/2^k} \ge \frac{k^2}{4.2^k}}.(1-2^{-k^2}) \le \frac{8}{k^2} $$
Thus $\sum_{n} a_n < \infty$
However , by definition
$$\limsup na_n=1$$
Discussion:

*

*There must be some kind of additional condition in order for that to be true.

*As shown in many previous topics, the condition of monotonicity is a pertinent choice. To me, that's conditional is somehow pretty tight...

*As I have shown in my counterexample above: a condition on the convergence rate of $ \limsup \frac{a_{n+1}}{a_n} =1$ is  also crucial.

