Suppose $x_{n}>0$ and tends to zero.We know its arithmetic mean tends to zero. I wonder which will tend to zero faster. In detail, $x_{n}>0$ and tends to zero, $S_{n}=\sum_{k=1}^{n}x_{k}$ (so 
 $\lim_{n\to\infty}\frac{Sn}{n}=0$). Let $T_{n}=\frac{nx_{n}}{S_{n}}.$ I think the "average" $S_n/n$ will tend to zero more slowly than $x_n$. In fact,I tried some simple examples,the limit of $T_{n}$is always zero.
So I wonder can the limit of $T_{n}$ be $\infty$? Or can the upper limit of $T_{n}$ be $\infty$?
 A: Warmup: non-zero limit.
Suppose $(x_n)_n$ is defined by
$$
x_n = \begin{cases}
\frac{1}{n} &\text{ if } n \text{ is a power of two}\\
\frac{1}{2^{2^{2^n}}} &\text{ otherwise.}
\end{cases}
$$
(The second condition is just to ensure positivity, for all intend and purpose we can (and will below) proceed with it at $0$).
Then
$$
S_{2^k} = \sum_{\ell=0}^k \frac{1}{2^\ell} = 2(1-1/2^{k+1})
$$
so 
$$
T_{2^k} = \frac{2^{k}x_{2^k}}{S_{2^k}} = \frac{1}{2(1-1/2^{k+1})} \xrightarrow[n\to\infty]{} \frac{1}{2} \neq 0
$$

Real deal: $\infty$ limit.
Similarly, suppose $(x_n)_n$ is defined by
$$
x_n = \begin{cases}
\frac{1}{\sqrt{n}} &\text{ if } n \text{ is a power of two}\\
0 &\text{ otherwise.}
\end{cases}
$$
(same thing as before, one can make that positive by replacing $0$ by an insane inverse tower of exponentials, for instance).
Then
$$
S_{2^k} = \sum_{\ell=0}^k \frac{1}{2^{\ell/2}} = \frac{1-1/2^{(k+1)/2}}{1-1/\sqrt{2}}
$$
so 
$$
T_{2^k} = \frac{2^{k}x_{2^k}}{S_{2^k}} = (1-1/\sqrt{2})\cdot \frac{2^{k/2}}{1-1/2^{(k+1)/2}} \xrightarrow[n\to\infty]{} \infty
$$
