Cesàro summability and $\sum n \lvert a_n\rvert ^2 < \infty$ implies convergence

Question

How can I prove that if $\sum_{n=1}^\infty a_n$ is Cesàro summable and if $\sum_{n=1}^\infty n |a_n|^2 < \infty$, then $\sum_{n=1}^\infty a_n$ converges?

Answer 1

Take $$ a_n=\frac{1}{n(\log(n+1))}. $$ Then $a_n\to 0$, and hence it is Cesaro summable. Also $$ \sum_{n=1}^\infty na_n^2=\sum_{n=1}^\infty\frac{1}{n(\log(n+1))^2}<\infty, $$ but $$ \sum_{n=1}^\infty \frac{1}{n(\log(n+1))}=\infty! $$

Answer 2

@Yiorgos S. Smyrlis I think your counter-example does not hold:

You can easily prove that if $S_n =\sum_{k=1}^n a_k $ :

$S_n$ ~ $\ln(\ln(n))$

Both being terms of diverging series you can say that partial sums are equivalent:

$ \sum_{k=1}^N S_k $ ~ $\sum_{k=1}^N \ln(\ln(k))$ ~ $\int_A^N \ln(\ln(t)) dt$

What we're trying to know is whether $\frac{1}{N}\sum_{k=1}^N S_k$ converges or not ( $\sum a_n$ Cesaro summable):

Let $t=Nu$ :

$B_n = \frac{1}{N}\int_A^N \ln(\ln(t)) dt =\int_{\frac{A}{N}}^1 \ln(\ln(Nu)) du =\int_{\frac{A}{N}}^1 \ln(\ln(N) +\ln(u)) du $

$ B_n = (1-\frac{A}{N})\ln(\ln(N)) + \int_{\frac{A}{N}}^1 \ln(1 +\frac{\ln(u)}{\ln(N)}) du$

$\ln(1+x) = x+ o(x)$

$\implies B_n =(1-\frac{A}{N})\ln(\ln(N)) + \frac{1}{\ln(N)}\int_{\frac{A}{N}}^1\ln(u) du + o(\frac{1}{\ln(N)})$

$\implies B_n=(1-\frac{A}{N})\ln(\ln(N))+K\frac{1}{\ln(N)}+o(\frac{1}{\ln(N)}) $

Finally that proves : $ \frac{1}{N}\sum_{k=1}^N S_k $ ~ $B_n$ ~ $\ln(\ln(N))$

So $\sum a_n$ is not Cesaro summable. I'll come back on the original question in I find something..

Edit Hello, it's been a while and I don't know if you found an answer already but I have something:

Let: $S_n =\sum_{k=1}^n a_k $ ; $C_n =\frac{1}{n}\sum_{k=1}^n S_k $

Then: $C_n =\sum_{k=1}^n (1-\frac{k}{n})a_k= S_n - \frac{1}{n}\sum_{k=1}^n ka_k $ Now we know that $(C_n)$ converges, and we are left with the right hand side to handle.

Let's use the second hypothesis, and the well-known result: $u_n\rightarrow u $ => $\frac{1}{n}\sum_{k=1}^n u_k\rightarrow u$

Let $H_n = \sum_{k=1}^n k|a_k|²$; $K_n = \frac{1}{n}\sum_{k=1}^n H_k$

$K_n = \sum_{k=1}^n (1-\frac{k}{n})k|a_k|² = H_n - \frac{1}{n}\sum_{k=1}^n k²|a_k|² $

Now because of the result I just mentioned, $(K_n)$ and $(H_n)$ have the same limit, hence: $\frac{1}{n}\sum_{k=1}^n k²|a_k|² \rightarrow 0$

$|\frac{1}{n}\sum_{k=1}^n ka_k| \leq \frac{1}{n}\sum_{k=1}^n 1*k|a_k| \leq \frac{1}{n} \sqrt{\sum_{k=1}^n 1²}\sqrt{\sum_{k=1}^n k²|a_k|²}=\sqrt{\frac{1}{n}\sum_{k=1}^n k²|a_k|²}$

So we get: $\frac{1}{n}\sum_{k=1}^n ka_k \rightarrow 0$

And from : $S_n= C_n + \frac{1}{n}\sum_{k=1}^n ka_k $ we get:

$S_n\rightarrow lim(C_n)$ : $(S_n)$ converges, CQFD