Let $(a_n)$ be a sequence such that $\liminf{|a_n|}=0$. Show that there exists a subsequence $a_{n_k}$ of $a_n$ such that $\sum_{k=1}^{\infty}a_{n_k}$ converges.
I want to make sure that my proof is correct for this, as it differs from the solution given by my professor. The solution he gave was far more elaborate, to the point at which it made me doubt whether or not mine was valid. This is my argument:
If $\liminf{|a_n|}=0$, then there must exist a subsequence of $|a_n|$, let's call it $|a_{n_k}|$, such that $|a_{n_k}|\rightarrow 0$. Then from the definition of limit convergence, we have that:
$$\forall \epsilon>0, \exists N:k>N\Rightarrow |a_{n_k}|<\epsilon$$
Then I tried to attack the problem using the cauchy criterion. Consider the summation:
$$\bigg|\sum_{k=m}^{n}|a_{n_k}|\bigg|$$
for $n,m>N$. We now know that:
$$\bigg|\sum_{k=m}^{n}|a_{n_k}|\bigg|\leq \sum_{k=m}^{n}|a_{n_k}|<(n-m)\epsilon$$
We want to show that:
$$\bigg|\sum_{k=m}^{n}|a_{n_k}|\bigg|<\epsilon'$$
for $n,m>N$ and $\epsilon'>0$, but as $\epsilon$ is arbitrary we can simply set $\epsilon = \frac{\epsilon'}{n-m}$. Then we end up with:
$$\bigg|\sum_{k=m}^{n}|a_{n_k}|\bigg|<(n-m)\frac{\epsilon'}{n-m}$$
Which would imply that the cauchy criterion holds for the subsequence $a_{n_k}$. Is this proof correct? Are there any suggestions on how I can improve it?
Best Answer
Your proof is not correct.
The issue is that $n,m$ depend on $\epsilon$. Hence you can't simply set $\epsilon = \frac{\epsilon^\prime}{n-m}$.
A way to tackle the problem is to build a subsequence such that $\lvert a_{n_k} \rvert \le \frac{1}{2^k}$ for all $k \in \mathbb N$.