If the average of a sequence converges, does the average of the square roots converge?

Question

I'm looking for a nice proof or counterexample for the following claim:

If $d_i \in [0,1]$ are such that $\lim_{n \to \infty} (d_1+...+d_n)/n$ is well-defined, then $\lim_{n \to \infty} (\sqrt{d_1}+...+\sqrt{d_n})/n$ is well-defined.

Answer 1

For $(2k)! \leqslant n < (2k+1)!$, let $d_n = \frac{1}{2}$, and for $(2k+1)! \leqslant n < (2k+2)!$ let $d_n = \frac{1}{2}\bigl(1 + (-1)^n\bigr)$.

Then we have

$$\lim_{n\to \infty} \frac{1}{n}\sum_{m = 1}^n d_m = \frac{1}{2},$$

but

\begin{align} \frac{1}{(2k)!}\sum_{m = 1}^{(2k)!} \sqrt{d_m} &\leqslant \frac{(2k-1)! + \frac{2k-1}{2}(2k-1)!}{(2k)!},\\ \frac{1}{(2k+1)!}\sum_{m = 1}^{(2k+1)!} \sqrt{d_m} &\geqslant \frac{\frac{1}{\sqrt{2}}2k(2k)!}{(2k+1)!}, \end{align}

so

$$\liminf_{n\to\infty} \frac{1}{n}\sum_{m = 1}^n \sqrt{d_m} \leqslant \frac{1}{2} < \frac{1}{\sqrt{2}} \leqslant \limsup_{n\to\infty} \frac{1}{n}\sum_{m = 1}^n \sqrt{d_m}.$$