3
$\begingroup$

Let $(a_n)_{n\in\mathbb{N}}\subseteq \mathbb{R}_{>0}$ be a convergent sequence. Is $$\sum\limits_{n=1}^\infty\frac{1}{n^2}(a_1+\cdots+ a_n)$$ convergent?

I don't know if this is correct or not and tried to find a counterexample so far, unsuccessfully, but I think it doesn't converge in general.

If one can choose $(a_n)_n$ as in this question Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively. , then $\sum\limits_{n=1}^\infty\frac{1}{n^2}(a_1+\cdots+ a_n)$ does not converge due to the divergence of the harmonic series. But I would like to have a more concrete counterexample.

I am happy about any hint and help. Thank you

$\endgroup$
5
$\begingroup$

Not at all just consider take $a_n = 1$ then

$$\sum_{n=1}^{\infty}\frac{1}{n^2}(a_1+\cdots+ a_n)=\sum_{n=1}^{\infty}\frac{1}{n} =\infty$$

Generally if we assume that $\lim_{n\to\infty}a_n= \ell\neq 0$ by Cesaro summation we hat $$\lim_{n\to\infty}a_n = \ell =\lim_{n\to\infty}\frac{a_1+\cdots+ a_n}{n}$$

Hence for $N$ large enough we have, $$\frac{a_1+\cdots+ a_n}{n}\ge \frac{\ell}{2}~~~~~~for ~~~n\ge N$$

that is $$\sum_{n=N}^{\infty}\frac{1}{n^2}(a_1+\cdots+ a_n)=\frac{\ell}{2}\sum_{n=N}^{\infty}\frac{1}{n} =\infty$$

$\endgroup$
  • $\begingroup$ yes... I was kind of blind. Thank you $\endgroup$ – user472520 Jan 19 '18 at 17:47
2
$\begingroup$

On the contrary, if $\lim_{n\to +\infty}a_n=C\neq 0$, then $\sum_{n\geq 1}\frac{A_n}{n^2}$ is surely divergent (here $A_n$ stands for $a_1+a_2+\ldots+a_n$, of course). Indeed $\lim_{n\to +\infty}\frac{A_n}{n}=C$ by Cesàro theorem, so $\sum_{n\geq 1}\frac{A_n}{n^2}$ is divergent by asymptotic comparison with $\sum_{n\geq 1}\frac{C}{n}$.

$\endgroup$
2
$\begingroup$

Even when $a_n\to 0$ this may fail. Let $b_n=a_1+...+a_n.$

Let $a_n=1/\log (n+1).$ Then $a_n>\int_{n+1}^{n+2}(1/\log x)dx.$

So for $n\geq 2$ we have

$$\begin{split} b_n &> \int_2^{n+2}\frac{1}{\log x}dx \\ &\geq \int_{(n+2)/2}^{n+2}\frac{1}{\log x}dx\\ &\geq \int_{(n+2)/2}^{n+2} \frac{1}{\log (n+2)}dx\\ &= \frac {n+2}{2\log (n+2)}. \end{split}$$

So for $n\geq 2$ we have $$b_n/n^2>\frac {n+2}{2n^2\log (n+2)}\geq\frac {1}{2(n+2)\log (n+2)}.$$ By the Cauchy Condensation Test $\sum_{n>1}\frac {1}{n\log n}$ diverges to $\infty .$ So $\sum_nb_n/n^2$ diverges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy