If $\lim{a_n}=0$ then there exists $a_{k_n}$ such that $\lim n a_{k_n}=0$ in $\mathbb R$ If $\lim{a_n}=0$    then there exists $a_{k_n}$ such that $\lim n a_{k_n}=0$  in $\mathbb R$
Attempt.
First I wrongly thought and divided given $\epsilon$ into $n$ parts and then sum. But it is wrong because $ n $ is not constant.
Second I tried to play with partial sums and their difference for example $2n a_n<\sum S_{2n}-S_n....$ but I realized I didnot have if it is convergence  or divergence series with $a_n$
Third I consider some kind of subsequences that analog to for example Cauchy s $2^n$ series... or consider the other subsequence functions but I failed..
How can I start? Can you give me a hint please?
 A: Given that $\lim_{n\to\infty}a_n=0$, there exists some index $k_1$ with $|a_{k_1}|<\frac1{1^2}$. Then there exists an index $k_2>k_1$ with $|a_{k_2}|<\frac1{2^2}$. We continue this process, forming an increasing sequence of indices $k_n$ with $|a_{k_n}|\le\frac1{n^2}$. Then
$$|na_{k_n}|=n|a_{k_n}|<\frac n{n^2}=\frac1n$$
and we conclude that $\lim_{n\to\infty}na_{k_n}=0$ by the squeeze theorem.
A: The other answers have answered your question (which amounts to asking: why, if $a_n\to0$, must there be a subsequence that vanishes faster than $n^{-1}$?). 
But there is an obvious generalization, which shows there is nothing special about the factor $n$ in your problem. Informally, the point is that if $a_n$ tends to zero, we can always select a subsequence that vanishes "as fast as we want" -- faster than the reciprocal of any sequence that diverges to infinity. The proof idea is exactly the same.

Claim. Suppose $a_n\to 0$ and $b_n\to\infty$ with $b_n\neq 0$ for all $n$. Then there exists a subsequence $a_{k_n}$ such that $b_na_{k_n}\to0$.

Proof. Define
$$k_n=\min\left\{i\,|\,i\in\mathbb{N}, |a_i|<b_n^{-2}\right\}$$
This definition is well-defined because $a_n\to0$ (and every set of natural numbers has a least element).
Then
$$0\leq|b_na_{k_n}|<|b_n|b_n^{-2}=|b_n^{-1}|$$
The result follows by the squeeze theorem.
Note that squaring is not essential. We could use $b_n^{-\alpha}$ for $\alpha>1$.
A: Hint: If $a_n\to 0$ then, for any $N$, there exists some $k_N$ so that, say,
$$\left|a_{k_N}\right|<\frac{1}{N^2}.$$
