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.

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 $n$.
