# 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 ?

• @Buraian Can you give a full proof please? Commented Nov 12, 2020 at 14:07
• Cauchy's condensation works only if $(a_n)$ is non-increasing. – With that additional condition, a proof is e.g. here: math.stackexchange.com/q/4603/42969 or here: math.stackexchange.com/q/1983651/42969. Commented Nov 12, 2020 at 14:22
• @MartinR Thanks for your correction. I forget that. We are not given the "non-increasing" condition, indeed. Commented Nov 12, 2020 at 14:30
• Where does the problem come from? Commented Nov 12, 2020 at 18:43

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.
• Can you give a simpler counterexample？ Commented Nov 13, 2020 at 7:14
• Perhaps there is a simpler one but it'll take time, and I already spent not so little time working on this problem. Commented Nov 13, 2020 at 11:59
• I also did a survey on all presented solutions in the previous topics. The methods used in those solutions can be reused to relax the monotonicity. That is, if we have: $$\limsup \left( \frac{a_{n+1}}{n} \right)^n <e$$ In instead of the monotonicity, we are still able to imply the limit $$\lim na_n =0$$ Remark 1 : This is a condtion of type convergence rate.<br> Remark 2 : The inequality is strict.<br> Remark 3 : IMHO, this condition is already tight. After all my work, I feel that I can provide a counterexample when that inequality is not strict. Commented Nov 13, 2020 at 14:21