This is a homework question

Prove if $\lim\limits_{n\rightarrow\infty}(a_1+a_2+...+a_n)=S$, then $\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+na_n}{n}=0$

And this is my try to prove it,

Let $k < n$,


There are k zeros, and they are finite. So $\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+ka_k}{n}=0$

Then, let $k\rightarrow\infty$, we get $$\begin{align*}\lim\limits_{k\rightarrow\infty}\left(\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+ka_k}{n}\right)&=\lim\limits_{k\rightarrow\infty}0\\&=0\end{align*}$$

which equals to $$\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+na_n}{n}=0$$

I'm not quite sure it's a correct proof. Especially, the step of letting $k < n$, I can't say that on what degree k is to be smaller than n. Since $n\rightarrow\infty$, then $k$ could also approach to $\infty$.

If it's not a valid proof, where goes wrong?

And I believe there are some better methods to prove it, please give me some hints.

Any suggestion will be appreciated.

  • $\begingroup$ is this graded homework? $\endgroup$ – Zelos Malum Oct 21 '15 at 3:49
  • $\begingroup$ No, it's just for practice. $\endgroup$ – 0xBBC Oct 21 '15 at 3:51
  • $\begingroup$ ah good, I would say your method doesn't work because you're trying to split it up in an unnatural manner where it seems more forced than logical. $\endgroup$ – Zelos Malum Oct 21 '15 at 3:53
  • $\begingroup$ Yes, I also got this feeling when I wrote it. If let $S_n = \sum\limits_{n\rightarrow\infty}a_n$, then rewrite the equation as $\lim\limits_{n\rightarrow\infty}\frac{nS_n-S_1-S_2-...-S_{n-1}}{n}$, so it's equal to $S_n-\lim\limits_{n\rightarrow\infty}\frac{S_1+S_2+...+S_{n-1}}{n}$. Can we say $S_n=\lim\limits_{n\rightarrow\infty}S_{n-1}=0$? $\endgroup$ – 0xBBC Oct 21 '15 at 4:01
  • $\begingroup$ The problem comes in with the last statement $$\lim\limits_{k\rightarrow\infty}\left(\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+ka_k}{n}\right)$$ is not necessarily equal to $$\lim\limits_{n\rightarrow\infty}\frac{a_1+2a_2+...+na_n}{n}$$ $\endgroup$ – Paul Sinclair Oct 21 '15 at 4:04

Well this is obvious if one uses the following result

Theorem: If $u_{n} \to L$ as $n \to \infty$ then $$\lim_{n \to \infty}\frac{u_{1} + u_{2} + \cdots + u_{n}}{n} = L$$

For the current question let $$S_{n} = a_{1} + a_{2} + \cdots + a_{n}$$ and we are given that $S_{n} \to S$ as $n \to \infty$. We have clearly $$a_{n} = S_{n} - S_{n - 1}, n > 1, a_{1} = S_{1}$$ and therefore \begin{align} b_{n} &= \frac{a_{1} + 2a_{2} + \cdots + na_{n}}{n}\notag\\ &= \frac{S_{1} + 2(S_{2} - S_{1}) + \cdots + n(S_{n} - S_{n - 1})}{n}\notag\\ &= \frac{nS_{n} - (S_{1} + S_{2} + \cdots + S_{n - 1})}{n}\notag\\ &= \frac{(n + 1)S_{n} - (S_{1} + S_{2} + \cdots + S_{n})}{n}\notag\\ &= \frac{n + 1}{n}\cdot S_{n} - \frac{S_{1} + S_{2} + \cdots + S_{n}}{n}\notag\\ &\to S - S = 0\text{ as }n \to \infty\notag \end{align}

Note: The theorem mentioned in the beginning is not obvious and its proof (although not too difficult) is available on MSE.

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  • $\begingroup$ Oh, I get it. Thank you for your great help! $\endgroup$ – 0xBBC Oct 21 '15 at 5:15

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