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(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}

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

3 Answers 3

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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.

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    $\begingroup$ (+1) All to easy. $\endgroup$
    – Mark Viola
    Aug 8, 2020 at 21:24
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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 $$ \frac{s_N}{N^2} \le A + B .$$

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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.

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