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I'm looking for slick proofs that if $a_n$ is a sequence of complex numbers such that $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$.

My not so slick proof:

Let $A_x=\sum\limits_{k=1}^{x} a_k$ and apply an Abel sum with the function $f(x)=\frac{1}{x}$. We get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\int\limits_1^n\frac{A_x}{x^2}dx=\frac{A_n}{n}.$$ Of course we can split the integral into chunks to get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{i=1}^n \left( \frac{1}{k+1}-\frac{1}{k} \right) A_k=\frac{A_n}{n}.$$ If we expand $A_k$ on the left side, it telescopes and yields $$\sum\limits_{k=1}^n \frac{a_k}{k} + \sum\limits_{k=1}^n \left( 1-\frac{1}{k+1}\right) a_k=\frac{A_n}{n}\iff \frac{(n-1)A_n}{n}=\sum\limits_{k=1}^n\frac{a_k}{k(k+1)}.$$ The norm of the right side of the equation is clearly $\mathcal O(\log(n))$ (using the triangle inequality and the fact that $\frac{a_k}{k}$ is bounded). The desired result follows after dividing both sides by $n-1$.

I would also appreciate some proof verification. Initially, I thought that the problem was going to be really easy, but it took me a bit of effort. Am I missing something major? The original problem is for real sequences, but I don't think this helps. Obviously if the sequence converged absolutely then it would also be absolutely trivial.

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  • $\begingroup$ Your notation is confusing me just a bit, so I want to make sure I have this right: in the first summation, $n\in \mathbb{N}$ and is fixed, but in the second summation we are talking about a different $n$ which is not fixed? $\endgroup$ – Brevan Ellefsen Jan 7 '17 at 18:36
  • $\begingroup$ which summations? You mean $\sum\limits_{k=1}^n \frac{a_k}{k}$ and $\lim\limits_{n\to\infty}\frac{1}{n} \sum\limits_{k=1}^n a_k$ ? $\endgroup$ – Jorge Fernández Hidalgo Jan 7 '17 at 18:38
  • $\begingroup$ Yes, those were the two I was referring to. $\endgroup$ – Brevan Ellefsen Jan 7 '17 at 18:39
  • $\begingroup$ I mean that $\lim\limits _{n\to \infty} \sum\limits_{k=1}^n \frac{a_k}{k}$ exists, and I want to prove $\lim\limits_{n\to \infty} \frac{1}{n} \sum\limits_{k=1}^n a_k$ is $0$. $\endgroup$ – Jorge Fernández Hidalgo Jan 7 '17 at 18:41
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    $\begingroup$ OH, ok! that makes a lot more sense. Thanks for the clarification. Not explicitly writing the limit on $\sum\limits_{k=1}^n \frac{a_k}{k}$ threw me off. $\endgroup$ – Brevan Ellefsen Jan 7 '17 at 18:43
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Let $b_k = \dfrac{a_k}{k}$, and define

$$R_m = \sum_{k = m}^{\infty} b_k.$$

Then

\begin{align} \frac{1}{n} \sum_{k = 1}^n a_k &= \frac{1}{n}\sum_{k = 1}^n kb_k \\ &= \frac{1}{n} \sum_{k = 1}^n\sum_{m = k}^n b_m \\ &= \frac{1}{n} \sum_{k = 1}^n \bigl(R_k - R_{n+1}\bigr) \\ &= \Biggl(\frac{1}{n} \sum_{k = 1}^n R_k\Biggr) - R_{n+1}. \end{align}

Now use that $R_k \to 0$, and that the Cesàro means of a convergent sequence converge to the same limit.

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  • $\begingroup$ Beautiful, I tried using Cezaro-Stolz but I couldn't do it. Thank you for this. I didn't think of using the tails of the series, although in retrospect it is a very natural way to use the convergence of the series and not just the fact that the terms go to $0$. $\endgroup$ – Jorge Fernández Hidalgo Jan 7 '17 at 18:53
  • $\begingroup$ Very elegant! ${}$ $\endgroup$ – user384138 Jan 7 '17 at 19:06
  • $\begingroup$ Why can we not use comparison? $\endgroup$ – Guacho Perez Jan 16 '17 at 15:32
  • $\begingroup$ @GuachoPerez Maybe one can. But what to compare it with? $\endgroup$ – Daniel Fischer Jan 16 '17 at 15:35
  • $\begingroup$ Can one use the fact that $\frac{a_n}{N}\le \frac{a_n}{n}$ in each partial sum? $\endgroup$ – Guacho Perez Jan 16 '17 at 15:38

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