# Suppose $\sum_{n=1}^\infty \frac{a_n}{n^2}< \infty$. Prove that $\lim\limits_{n\to\infty} \frac{s_n}{n^2}= 0$.

(a) Let $$(a_n)^\infty_{n=1}$$ be a sequence of non-negative real numbers such that $$\sum_{n=1}^\infty \frac{a_n}{n^2}< \infty$$. Let $$(s_n)^\infty_{n=1}$$ be a sequence defined for all $$n \in \mathbb{N}$$ by $$s_n=\sum_{k=1}^n a_k$$. Prove that $$\lim\limits_{n\to\infty} \frac{s_n}{n^2}= 0$$.

(b)Use the sequence $$a_n =\frac{n}{(1 + \log n)}$$ to prove that the converse of part (a) does not hold in general.

I tried in the following way:

There is a theorem, If $$\sum a_n$$ converges then $$\lim\limits_{n\to\infty} a_n =0$$. I want to use the comparison test to show $$\sum s_n/n^2$$ converges.

$$\frac{s_n}{n^2}=\frac{\sum_{k=1}^n a_k}{n^2} \leq \sum_{n=1}^\infty \frac{a_k}{n^2} < \infty$$

So, $$\sum s_n/n^2$$ converges and hence, $$\lim\limits_{n\to\infty} \frac{s_n}{n^2}= 0$$.

On the other hand, for $$a_n=\frac{n}{1+\log n}$$, $$\sum \frac{a_n}{n^2}$$ diverges if $$\int_{1}^\infty \frac{a_n}{n^2}$$ diverges.

\begin{align} & \int_1^M \frac{a_n}{n^2}= \int_1^M \frac{1}{x(1+ \log x)} \, dx \\[6pt] = {} & \log u \Big\vert_1^{1+\log M}=\log(1+ \log M )-\log 1 \\[6pt] = {} & \log(1+\log M) \to \infty \text{ as } M \to \infty. \end{align}

• $s_n = \sum^n a_k$ ? you should perhpas write it down Aug 8, 2020 at 18:51
• In general $\sum s_n/n^2$ doesn't converge. Take $a_n = 1$ for all $n$, then $s_n = n$. Aug 8, 2020 at 18:55
• @DanielFischer the limit of $s_n / n^2 = n/n^2=1/n$ is zero. Aug 8, 2020 at 18:57
• @CyclotomicField Yes. Since $\sum a_n/n^2$ converges, we have $s_n/n^2 \to 0$. But $\sum s_n/n^2$ does not converge, contrary to what the OP asserts. Aug 8, 2020 at 18:59
• @CyclotomicField I doubt that, since it's explicitly stated that the comparison test is to be used to show convergence of $\sum s_n/n^2$. Aug 8, 2020 at 19:08

For any $$\varepsilon>0$$, there is $$N$$ such that $$0\leq \sum_{n> N}\frac{a_n}{n^2}\leq \varepsilon$$ For all such $$n$$, $$\frac{s_n}{n^2}=\frac{s_N}{n^2}+\frac{\sum^n_{m>N}a_m}{n^2}\leq \frac{s_N}{n^2}+\sum^n_{m>N}\frac{a_m}{m^2}<\frac{s_N}{n^2}+\varepsilon$$

Letting $$n\rightarrow\infty$$ gives

$$\limsup_n\frac{s_n}{n^2}\leq\varepsilon$$ for all $$\varepsilon>0$$. The conclusion follows from here.

• (+1) All to easy. Aug 8, 2020 at 21:24

Given $$\epsilon>0$$, there exists $$N_1$$ such that $$A = \sum_{n=N_1+1} ^ \infty \frac{a_n}{n^2} < \epsilon .$$ Then there exists $$N_2$$ such that $$B = \frac1{N_2^2} \sum_{n=1}^{N_1} a_n < \epsilon .$$ And if $$N > \max\{N_1, N_2\}$$, then $$\frac{s_N}{N^2} \le A + B .$$

Proof. (a) For every $$\epsilon >0$$, there exists $$k$$ such that $$\sum_{m>k}\frac{a_m}{m^2}<\epsilon.$$ It follows that for any $$n>k,$$ one has $$\sum_{m=k+1}^n \frac{a_m}{n^2}\leq \sum_{m=k+1}^n\frac {a_m}{m^2}<\epsilon$$ $$\Rightarrow \frac{s_n-s_k}{n^2}<\epsilon$$ $$\Rightarrow \limsup_{n\rightarrow \infty}\frac {s_n}{n^2}\leq \lim_{n\rightarrow \infty}\left(\frac {s_k}{n^2}+\epsilon\right)=\epsilon.$$ Since $$\epsilon$$ is arbitrary, one has $$\lim_{n\rightarrow\infty}\frac{s_n}{n^2}=0.$$

(b) To show that the converse is not true, you have checked that for $$a_n=\frac n{1+\log(n)}$$, $$\sum_{n=1}^\infty\frac{a_n}{n^2}$$ diverges. It remains to show that $$\lim_{n\rightarrow \infty}\frac {s_n}{n^2}=0.$$ Observe that $$\left(\frac x{1+\log(x)}\right)'=\frac{\log(x)}{(1+\log(x))^2}\geq 0,~{\rm for~}x\geq 1.$$ It follows that $$s_n=\sum_{m=1}^na_m\leq \int_1^{n+1}\frac x{1+\log(x)}~dx=\frac{N(n)}{1+\log(N(n))}\cdot n,\quad (1)$$ where $$1\leq N(n)\leq n+1$$ (by the mean value theorem for integral). One now shows that $$N(n)\rightarrow \infty$$ as $$n\rightarrow \infty.$$

Lemma 1. For $$x\geq 1,$$ one has $$\sqrt{2x}\geq 1+\log(x).$$

Proof. $$\sqrt{2x}\geq 1+\log(x)$$ $$\Leftrightarrow 2x\geq (1+\log(x))^2,$$ the latter is true if the function $$f(x):=2x-(1+\log(x))^2$$ satisfies $$f(1)\geq 0$$ and $$f'(x)\geq 0$$ for all $$x\geq 1.$$ Clearly $$f(1)=1>0$$ and $$f'(x)=2-2(1+\log(x))\cdot \frac 1 x.$$ Now $$f'(x)\geq 0$$ for $$x\geq 1$$ is equivalent to $$x\geq 1+\log(x)$$ for $$x\geq 1,$$ which is true since $$g(x):=x-1-\log(x)$$ satisfies $$g(1)\geq 0$$ and $$g'(x)\geq 0$$ for $$x\geq 1.$$

To prove the assertion that $$N(n)\rightarrow \infty$$ as $$n\rightarrow \infty,$$ one uses (1) and Lemma 1: $$\int_1^{n+1}\frac x{\sqrt{2x}}~dx\leq \int_1^{n+1}\frac x{1+\log(x)}~dx=\frac{N(n)}{1+\log(N(n))}\cdot n$$ $$\Rightarrow \frac{\sqrt{2}}3((n+1)^{3/2}-1)\leq \frac{N(n)}{1+\log(N(n))}\cdot n$$ $$\Rightarrow \frac{\sqrt{2}}3\cdot \frac{(n+1)+(n+1)^{1/2}+1}{(n+1)^{1/2}+1}\leq \frac{N(n)}{1+\log(N(n))}\leq N(n),$$ which shows that $$N(n)\rightarrow \infty$$ as $$n\rightarrow \infty.$$ Using this, one concludes from (1) that $$s_n\leq \frac{N(n)}{1+\log(N(n))}\cdot n\leq \frac{(n+1)n}{1+\log(N(n))}$$ $$\Rightarrow \frac {s_n}{n^2}\leq \frac 1{1+\log(N(n))}\cdot\frac{n+1}{n},$$ hence $$\lim_{n\rightarrow \infty}\frac {s_n}{n^2}=0,$$ as required.