11

We are given that $\displaystyle\sum_{k=1}^\infty \frac{a_k}{k}$ converges, and we want to show that $$\lim_{n\to \infty} \frac{a_1 +a_2 + \cdots + a_n}{n} = 0.$$ Let $\epsilon>0.$ Then since $\displaystyle\sum_{k=1}^\infty \frac{a_k}{k}$ converges we have that there is some positive integer $N_\epsilon$ such that $\left|\sum_{k=N_\epsilon}^\infty \dfrac{a_k}{k}\right|<\dfrac{\epsilon}2$. Let $\sum_{k=1}^{N_\epsilon} a_k =A$ and choose $N_1$ so that $\left|\frac{A}{n}\right|<\frac{\epsilon}{2}$ when $n\geq N_1$. Now let $n\geq \max \left\{N_1, N_\epsilon \right\}$ so that $$\left|\frac{a_1 +a_2 + \cdots + a_n}{n}\right|\leq\left|\frac{A}{n}\right|+\left|\frac{\sum_{k=N_\epsilon+1}^n a_k}{n}\right|<\left|\frac{A}{n}\right|+ \left|\sum_{k=N_\epsilon+1}^\infty \frac{a_k}{k}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$ Again, I do not feel convinced of my own argument. Am I going the wrong direction? Is there perhaps a better route to showing my conclusion? Any help is appreciated.

EDIT: Don’t know why this question got closed. I’m not solely wondering how to prove this statement, I’m also wondering if my proof is sufficient.

Chris Christopherson
  • 1,018
  • Yes, your argument doesn't work. You are using inequalities for $i$ between $1$ and $N_\epsilon$ to bound $\frac{a_i}{n}$ for $i$ that are not in that range. For example, $i=n-1$. – plop Jul 02 '21 at 21:19
  • @Urtur I don’t think I understand your critique. – Chris Christopherson Jul 02 '21 at 21:27
  • In the last displayed equation, in the second inequality, the second to last summand is $\left|\frac{a_{n-1}}{n}\right|$, which you bounded, I suppose, by $\frac{\epsilon}{2N_\epsilon}$. We only have the inequalities $\left|\frac{a_i}{n}\right|<\frac{\epsilon}{2N_\epsilon}$ for $1\leq i\leq N_\epsilon$. For $n$ large $n-1$ is not in that range. – plop Jul 02 '21 at 21:30
  • 2
    My thoughts: given $\epsilon\gt 0$, choose $M:\sum_{k=M}^\infty\frac {a_k}k\lt \frac{\epsilon}2$ (exists by series convergence). Separately, choose $N:\frac 1N\sum_{k=1}^Ma_k\lt\frac{\epsilon}2$ (similar reasoning). But connecting these two is an incomplete step at the moment... – abiessu Jul 02 '21 at 21:33
  • 1
    The easiest check is ... consider $a_n=(-1)^n$. Then $$\left|\frac{a_1}{n}\right|+\left|\frac{a_2}{n}\right|+\cdots + \left|\frac{a_n}{n}\right|=1$$ – rtybase Jul 02 '21 at 21:42
  • @Urtur ahhh I understand. Yes that’s where my argument fails. But we can pinch the entire tail of the series as abiessu uggested? And then combine that with my argument I think we have a solid proof no? – Chris Christopherson Jul 03 '21 at 01:46
  • @abiessu I think we can combine this idea with mine to complete this proof. See my reply to the first commenter. – Chris Christopherson Jul 03 '21 at 01:49
  • @Urtur I have edited my post accordingly. – Chris Christopherson Jul 03 '21 at 02:05
  • That still doesn't work. Instead of writing $...$, write explicit summation limits with $\sigma$ notation. That should help you note which terms are in each sum and which bound are you using for each. Also, four comments above you were given an explicit example with $a_n=(-1)^n$ that should help you realize that that bound doesn't work. – plop Jul 03 '21 at 12:40
  • @Urtur I don’t see why this choice for $a_n$ effects my argument. – Chris Christopherson Jul 03 '21 at 15:01
  • Well, to see how it affects your argument you need to be looking at your argument, which was what was written before you changed it into someone else's argument. – plop Jul 03 '21 at 20:54
  • @Urtur I don't think that this is someone else's argument. My argument has virtually remained the same this whole time. I have made it more concise. – Chris Christopherson Jul 06 '21 at 18:44

2 Answers2

15

For a "better route", use summation by parts with $S_0 = 0$ and $S_k = \sum_{j=1}^k \frac{a_j}{j}$ for $k > 0$ to get

$$\sum_{k=1}^n a_k = \sum_{k=1}^n k\frac{a_k}{k} = \sum_{k=1}^n k(S_k - S_{k-1}) = nS_n - \sum_{k=1}^{n-1}S_k(k+1 - k), $$

and, thus,

$$\frac{1}{n} \sum_{k=1}^n a_k = S_n - \frac{1}{n} \sum_{k=1}^{n-1}S_k$$

Finish by showing that the limit of the RHS is $0$. That $S_n \to S$ implies $n^{-1}\sum_{k=1}^n S_k \to S$ is a well known result.

RRL
  • 90,707
  • 1
    A beautiful answer : this is why i like maths.(+1). – hamam_Abdallah Jul 02 '21 at 22:00
  • I wish I could come up with stuff like this. Is there any way you can add a step between $$\sum_{k=1}^n k(S_k -S_{k-1})=nS_n-\sum_{k=1}^{n-1} S_k(k+1-k)?$$ When I write out the sum I end up at the same place as it has a bit of a “telescoping” nature to it. Also can you lead me to a proof of the “well known result” you refer to? Thank you so much! – Chris Christopherson Jul 03 '21 at 01:28
  • 1
    @ChrisChristopherson: The well known result is called Cauchy's first limit theorem. See the proof here – RRL Jul 03 '21 at 04:22
  • @RRL Hi, could I ask your assistance here, please? I will very grateful to you if you will help me because unfortunately it is some time I am trying to prove this without success and I am not really able find the solution. – Antonio Maria Di Mauro Jul 06 '21 at 18:14
  • @RRL Perhaps I found an answer but I do not really know if it is correct: so could you take a look at it? In particular I'd like to legitimate the use of fundamental theorem of Calculus and the interchange of integrals. So what can you say about? – Antonio Maria Di Mauro Jul 06 '21 at 20:21
4

I'll prove with your ideas.

Given that $\sum_{k=1}^{\infty}\frac {a_{k}}k$converges, for each $\epsilon$, there exist an $N_{\epsilon}$, such that$$|\sum_{k=N_\epsilon}^{\infty}\frac {a_k}k|<\frac \epsilon2$$for arbitrary $N'>N_\epsilon$, we have $$|\sum_{k=N_\epsilon}^{N'}\frac {a_k}{N'}|<|\sum_{k=N_\epsilon}^{N'}\frac {a_k}k|<\frac \epsilon2$$ while fix this $\epsilon$, $$\forall N_1>\frac {|2{\sum_{k=1}^{N_{}\epsilon}}{a_k}|}{\epsilon},|\frac {\sum_{k=1}^{N_{\epsilon}}{a_k}}{N_1}|<\frac \epsilon2$$ So, set N as max{$N_1,N_{\epsilon,}$}, we can get the conclusion$$\forall n>N,|\frac {\sum_{k=1}^{n}a_n}n|\leq |\frac {\sum_{k=1}^{N_{}\epsilon}{a_k}}n|+|\frac {\sum_{k=N_{}\epsilon}^{n}{a_k}}n|\leq \epsilon$$as the select of $\epsilon$ is arbitrary, the proof is complete.

Infinity_hunter
  • 5,369
  • 1
  • 10
  • 30
Barterlog
  • 163
  • Yes this is the idea. I see how my method is using too many “N”s, and is a bit overkill. – Chris Christopherson Jul 03 '21 at 14:39
  • How do you get the first inequality, if $\sum_{k=1}^\infty \frac{a_k}{k} = S$, where $S$ need not be $0$? It should be $\left\lvert \left(\sum_{k=1}^{N'} a_k/k\right) - S \right\rvert < \epsilon/2$, no? – V.S.e.H. Jul 30 '21 at 17:01