# If $\sum a_n$ converges then $\liminf na_n=0$

I am trying to prove the following statement:

Let $(a_n)_{n\in\mathbb N}$ be a positive sequence such that the series $\sum a_n$ is convergent. Prove that $$\liminf_{n}na_n=0$$

Now, if $(a_n)_{n\in \mathbb N}$ is a real valued sequence, then $\liminf a_n=\alpha$ if and only if the following two conditions hold:

1. If $\beta \in \mathbb R \cup \{\pm \infty\}$ is such that $\beta<\alpha$ there exists $n_0\in \mathbb N$ such that $a_n>\beta$ for every $n>n_0$
2. There exists a subsequence $(a_{\varphi (k)})_{k\in \mathbb N}$ of $(a_n)_{n\in \mathbb N}$ that converges to $\alpha$.

The first one is trivial since $(na_n)_{n\in \mathbb N}$ is positive, so the problem basically boils down to constructing a subsequence on $na_n$ that converges to zero.

The given hypothesis implies that $a_n\to 0$ and that $a_n<\frac{1}{n}$ eventually, but I cannot seem to be able to use these bits of information to construct a subsequence that converges to $0$. I also tried by contradiction, assuming that $\liminf na_n\neq 0$, but did not get very far. This brings me to the two following questions:

• Is it possible to explicitly construct a subsequence of $(na_n)_{n\in \mathbb N}$ that converges to zero?

• How can I prove the given statement?

• If $\alpha > 0$, then we have $a_n > \frac{\alpha}{2n}$ for all sufficiently large $n$. What can you conclude? – Sangchul Lee Jan 31 '17 at 17:23

## 1 Answer

A constructive argument is not always necessary - it is not, at least, in this problem; we shall reason by contradiction. Assume that $\liminf n a_n > 0$. This means that no subsequence of $(na_n)_n$ tends to $0$, which means that there exist $n_0 \in \Bbb N$ and $r>0$ such that $n a_n \ge r$ for $n \ge n_0$ (in simple words: from $n_0$ onwards no term of $(n a_n)_n$ comes closer to $0$ than $r$ - otherwise, if there were terms coming arbitrarily close to $0$, they would form a subsequence tending to $0$, which would contradict our assumption). In this case, then,

$$\sum _{n=0} ^\infty a_n = \sum _{n=0} ^{n_0 - 1} a_n + \sum _{n=n_0} ^\infty a_n \ge \sum _{n=0} ^{n_0 - 1} a_n + \sum _{n=n_0} ^\infty \frac r n = \infty$$

because the second term is essentially the harmonic series which is divergent - but this contradicts the convergence of $\sum a_n$! Therefore, our assumption must be false, so $\liminf n a_n = 0$.