2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ If $\alpha > 0$, then we have $a_n > \frac{\alpha}{2n}$ for all sufficiently large $n$. What can you conclude? $\endgroup$ – Sangchul Lee Jan 31 '17 at 17:23
6
$\begingroup$

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 $\liminf n a_n = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.