TRUE or FALSE: If a sequence converges to $0$, then it has a summable subsequence Let $\left( {{a_n}} \right)_{n = 1}^\infty $ be a sequence such that $\mathop {\lim }\limits_{n \to \infty } |{a_n}| = 0$. Prove that there are is subsequence ${a_{{n_k}}}$ so that $\sum\limits_{k = 1}^\infty  {{a_{{n_k}}}} $ converges.
Proof: Let $\left( {{a_n}} \right)_{n = 1}^\infty $ be a sequence and $\mathop {\lim }\limits_{n \to \infty } |{a_n}| = 0$ and ${a_{{n_k}}}$ be a subsequence.
Since $\lim\limits_{n \to \infty } |{a_n}| = 0$, let $\varepsilon  > 0$ , there is $N > 0$ so that 
    $$||a_n| - 0| < \frac{\varepsilon }{{m - i}},$$ that is $$|a_n| < \frac{\varepsilon }{{m - i}}$$
for all $n \geqslant N$ and $m,i \in N$ which $m > i.$
Consider subsequence ${a_{{n_k}}}$ , $k = i,i + 1,...,m - 1,m,[m - i{\text{ terms}}]$
Since $|{a_n}| < \frac{\varepsilon }{{m - i}}$ hence $$\begin{align}|{a_{{n_k}}}| &< \frac{\varepsilon }{{m - i}}|{a_{{n_{i + 1}}}} + {a_{{n_{i + 2}}}} + ... + {a_{{n_{m - 1}}}} + {a_{{n_m}}}|\\ &\leqslant |{a_{{n_{i + 1}}}}| + |{a_{{n_{i + 2}}}}| + ... + |{a_{{n_{m - 1}}}}| + |{a_{{n_m}}}|\\ &< \frac{\varepsilon }{{m - i}} + \frac{\varepsilon }{{m - i}} + ... + \frac{\varepsilon }{{m - i}}\\ &= \frac{{(m - i)\varepsilon }}{{m - i}}\\ &= \varepsilon\end{align}$$
$$\left|\sum_{k = i + 1}^m a_{n_k}\right|  < \varepsilon $$
By CAUCHY CRITERION FOR SERIES $\sum\limits_{k = 1}^\infty  {{a_{{n_k}}}} $ converges.
 A: Hint: For every $n$ and $i$ there exists $k\geqslant n+1$ such that $|a_k|\leqslant1/2^i$. Show this fact, then use it recursively to exhibit an increasing sequence $(\varphi(n))$ such that $|a_{\varphi(n)}|\leqslant1/2^n$ for every $n$.
A: If I am not mistaken, then your purported proof shows that every subsequence is summable, hence if it was correct it would show that every sequence converging to zero is summable. This is not the case.
A correct proof will start with a given sequence converging to zero and then show how to produce, using properties of that particular sequence, a subsequence which is summable, i.e. one which not only converges to zero but converges fast enough.
A: You seem to be trying to show that every subsequence (of a sequence converging to $0$) has a convergent series, since the subsequence you're looking at is arbitrary. This is not true, though. Consider the sequence $a_n=\frac1n$.
Instead, try to proceed as follows: Pick your favorite convergent series $\sum_{k=1}^\infty b_k,$ where all the $b_k$ are positive. Now, construct a subsequence $\{a_{n_k}\}_{k=1}^\infty$ of $\{a_n\}_{n=1}^\infty$ such that $|a_{n_k}|<b_k$ for all $k$. You'll need to be a bit careful in your choices, since we need $n_1<n_2<n_3<...,$ but the fact that $a_n\to 0$ will make it possible. Why is it that we can conclude that $$\sum_{k=1}^\infty a_{n_k}$$ is convergent, then?
