$a_n\to\infty$ then how does $\sum_{k=1}^{n} a_k$ diverge? Could anyone please give me some hint on proving or disproving the following:
If $a_n\to\infty$, then $\lim_{n\to\infty}\frac1{na_n}\sum_{k=1}^{n} a_k$ converges.
The examples I can think of all say that this sum in convergent. Is this statement then "really" always true?
 A: Without any further hypothesis, the statement is wrong: some values of $a_n$ can be way very small compared to the others.
Define $a_n = \begin{cases} k & \mbox{ if $n = 2k$ is even,} \\
2^k & \mbox{if $n = 2k+1$ is odd.}
\end{cases}$.
For $n = 2k$, splitting the sum between odd and even indices yields : $$\frac{1}{na_n}\sum \limits_{i=1}^n a_i = \frac{1}{2k^2} \Big(\sum \limits_{i=1}^k i + \sum \limits_{i=1}^{k-1} 2^i \Big) = \frac{1}{2k^2} \Big(\frac{k(k+1)}{2} + 2^k - 1\Big) \ge \frac{2^k}{2k^2}.$$
Similarly, for $n=2k+1$, we get $$\frac{1}{na_n}\sum \limits_{i=1}^n a_i = \frac{1}{(2k+1)2^k} \Big( \frac{k(k+1)}{2} + 2^{k+1}-1 \Big) = \frac{k(k+1)}{(4k+2) \cdot 2^k} + \frac{1}{2k+1} \cdot \frac{2^{k+1}-1}{2^k}.$$
Hence $\frac{1}{2ka_{2k}}\sum\limits_{i=1}^{2k}a_i \underset{k \to \infty}{\longrightarrow} +\infty$, while $\frac{1}{(2k+1)a_{2k+1}}\sum\limits_{i=1}^{2k+1}a_i \underset{k\to\infty}{\longrightarrow}0$. Thus $\big(\frac{1}{na_n}\sum\limits_{i=1}^na_i\big)_n$ does not converge.
