Suppose the sequence $\{a_n\}_{n=1}^{\infty}$ satisfies $$\mid\sum\limits^{n}_{k=1}{a_{k}}\mid\leq C\sqrt{n} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1, 2, 3, \cdots$$ where $C$ is a fixed (but arbitrary) number. Prove that the series $$\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$$ converges.

My attempt: Suppose $b_n:= \frac{1}{n}$; Abel's lemma on summation by parts gives $$\sum\limits^{k}_{n=1}{\frac{a_n}{n}}=\sum\limits^{k-1}_{n=1}{[\sum\limits_{i=1}^{n}{a_i}\cdot(b_n-b_{n+1})] + \sum\limits_{i=1}^{k}{a_i}\cdot b_k}$$

$$<\sum\limits^{k-1}_{n=1}{\mid\sum\limits_{i=1}^{n}{a_i}\mid\cdot(b_n-b_{n+1}) +\mid \sum\limits_{i=1}^{k}{a_i}\mid \cdot b_k}$$

$$\le\sum\limits_{n=1}^{k}[C\sqrt{n}\cdot{(\frac{1}{n}-\frac{1}{n+1}})]+C\sqrt{k}\cdot \frac{1}{k+1}$$


Since $k\rightarrow\infty$, therefore $$\frac{C\sqrt{k}}{k+1}\rightarrow 0.$$

Moreover, for the sigma notation, since $$\frac{C\sqrt{n}}{n(n+1)}<\frac{C\sqrt{n}}{n^2}=\frac{C}{n^\frac{3}{2}}$$

Above is a $p$-series with $p=\frac{3}{2}>1$, hence the series $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges. Even though $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges and $\frac{C\sqrt{k}}{k+1}$ approaches to $0$ for $k\rightarrow\infty$, but they do not imply the series $\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$ converges.

My proof seems not yet completed. How do I continue it?

  • $\begingroup$ You switch between $k$ and $n$ in your solution, you should fix that. $\endgroup$ – Jakobian Nov 22 '18 at 14:50
  • $\begingroup$ I had edited it. Thanks for pointing my mistake. $\endgroup$ – weilam06 Nov 22 '18 at 15:02

Let $$ s_n=\sum_{k=1}^na_n $$ then $|s_n|\le C\sqrt{n}$ and $$ \begin{align} \sum_{k=1}^n\frac{a_k}k &=\sum_{k=1}^n\frac{s_k-s_{k-1}}k\\ &=\sum_{k=1}^n\frac{s_k}k-\sum_{k=0}^{n-1}\frac{s_k}{k+1}\\ &=\sum_{k=1}^ns_k\left(\frac1k-\frac1{k+1}\right)+\frac{s_n}{n+1}\\ &=\sum_{k=1}^n\underbrace{\frac{s_k}{k(k+1)}}_{\le C\frac{\sqrt{k}}{k^2}}+\underbrace{\ \ \frac{s_n}{n+1}\ \ }_{\le C\frac{\sqrt{n}}{n\vphantom{k^2}}} \end{align} $$ The sum converges and the extra term vanishes.

Comment brought into the answer

Since it does clarify the answer, I will bring the following comment into the answer.

The sequence of partial sums $\sum\limits_{k=1}^n\frac{a_k}k$ is the same as the sequence $\sum\limits_{k=1}^n\frac{s_k}{k(k+1)}+\frac{s_n}{n+1}$ the latter sum converges absolutely, hence the partial sums are Cauchy and the extra term converges to $0$ hence it is Cauchy. The sum of two Cauchy sequences is Cauchy. Since the original sequence of partial sums is Cauchy, it converges.

  • $\begingroup$ You only prove an upper bound for $|\sum_{k=1}^n \frac{a_k}{k}|$. The right-way to prove the convergence is to show that $\sum_{k=1}^\infty \frac{a_k}{k}$ is a Cauchy-sequence! $\endgroup$ – p4sch Nov 22 '18 at 14:56
  • $\begingroup$ Not exactly. Robjohn showed that the series converges absolutely. $\endgroup$ – Jakobian Nov 22 '18 at 15:06
  • $\begingroup$ No, he didnt proved that the series converges absolutely! In general, it is false! Counterexample: $a_n = (-1)^n$! Problem: We dont have a bound for $\sum_{k=1}^n |a_k|$ only for $\sum_{k=1}^n a_k$. $\endgroup$ – p4sch Nov 22 '18 at 15:07
  • $\begingroup$ @p4sch: the series in the last line converges absolutely by comparison to $\sum k^{-3/2}$. That that series converges shows that the sequence of partial sums is Cauchy, and the series of partial sums are the same, $\endgroup$ – robjohn Nov 22 '18 at 15:10
  • $\begingroup$ As I already said, you only prove that the sequence $(\sum_{k=1}^n a_k/k)_{n \in \mathbb{N}}$ is bounded. Of course, this argument can be adapted in order to prove that this sequence is Cauchy! However, your answer doesn't prove convergence! $\endgroup$ – p4sch Nov 22 '18 at 15:13

In order to prove the convergence, it is enough to show that this sequence is a Cauchy-sequence (another way to prove the convergence is given by robjohn), because $\mathbb{R}$, resp. $\mathbb{C}$ is complete. Write $A_k := \sum_{i=1}^k a_i$ and $B_k := \sum_{i=1}^k a_i/i$. Note that $a_k = A_{k} - A_{k-1}$ Then for $n>m$ \begin{align} |B_n - B_m| = \left| \sum_{i=m+1}^n \frac{a_i}{i} \right| &= \left| \sum_{i=m+1}^n \frac{A_i -A_{i-1}}{i} \right| \\ & = \frac{|A_m|}{m+1} + \frac{|A_n|}{n} + \left| \sum_{i=m+1}^{n-1} A_i \left(\frac{1}{i} - \frac{1}{i-1} \right) \right| \end{align} Since $(i^{-1}-(i-1)^{-1})= -(i(i-1))^{-1}$ Now we can use the bound to get the bound $$\frac{2C}{\sqrt{m}}+ C \sum_{i=m+1}^{n-1} \frac{1}{(i-1) \sqrt{i}}.$$ The last sum can be estimated by $\int_{m-1}^\infty \frac{1}{x^{3/2}} d x < \frac{3}{2} (m-1)^{-1/2}.$ All in all, we see that $$|B_n-B_m| \le \frac{3C}{(m-1)^{1/2}},$$ i.e. this is a Cauchy-sequence as claimed.

Note that this argument works also if we only require that $$|\sum_{k=1}^n a_k| \le C n^{1-\varepsilon}.$$ Moreover, we cannot expect that $\sum_{k=1}^\infty a_k/k$ converges absolutely. Just take $a_k = (-1)^k$, then $$|\sum_{k=1}^n a_k| \le 1 \le \sqrt{n},$$ but $$\sum_{k=1}^\infty \frac{|a_k|}{k} = \sum_{k=1}^\infty \frac{1}{k} =\infty.$$


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