# Limit of $na_n$

This has been answered before, but I have another "solution" that looks suspicious.

Exercise 22 from Spivak's Calculus, Chapter 23: Suppose $\{a_n\}$ is decreasing and each $a_n \ge 0$ . Prove that if $\sum_{n=1}^\infty a_n$ converges, then $\lim _{n \to \infty} na_n =0$

If I use $$na_n \lt 1+na_n \le (1+a_n)^n$$ (from the binomial expansion) then $$\limsup_{n \to \infty} \sqrt[n]{na_n} \lt \limsup_{n \to \infty}(1+a_n)=1$$ since $\sum a_n$ converges thus $a_n \to 0$

By the root test $\sum na_n$ is convergent and so $\lim _{n \to \infty} na_n =0$

But I haven't used that $\{a_n\}$ is decreasing

• Follow the hint given in the problem: "Write down the Cauchy criterion and be sure to use the fact that $\{a_n\}$ is decreasing" – lhf Feb 23 '16 at 2:42
• – Martin Sleziak Mar 31 '16 at 18:31

In the second displayed formula, the one with the $\limsup$, we should have $\le$, not $\lt$. And the $\limsup$ being $1$ is not useful.
Under the given hypotheses, it is not necessarily true that $\sum na_n$ converges. For example, let $a_n=\frac{1}{n^2}$ or let $a_n=\frac{1}{n\sqrt{n}}$.
We know that $A_n=\sum_{k=1}^{n}{a_k}$ is a Cauchy sequence as $\sum{a_n}$ converges. Thus, for every $\epsilon>0$, exists $n_0$ such that for $n>n_0$ it holds $\epsilon>\sum_{k=1}^{n}{a_k}-\sum_{k=1}^{n/2-1}{a_k}\geq\frac{n}{2}a_n$ and then $$\frac{n}{2}a_n<\epsilon \Rightarrow na_n <2\epsilon$$
and we get that $\lim_{n\to \infty}{na_n}=0$