Let $a_n$ be a convergent sequence who's limit is $L$ (Should also apply to when its limit is infinite). Let $t_n$ be a sequence of positive numbers such that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{t_n}} = \infty $. Let $b_n$ be as follows: $${b_n} = {{\sum\limits_{k = 1}^n {{t_n}{a_n}} } \over {\sum\limits_{k = 1}^n {{t_n}} }}$$

Show that $b_n\to L$.

I got a hint that the proof should be similar to the regular Cesaro mean proof, but I wasn't able to pin in down in the finite case, not to mention the infinite one (I wasn't able to proof the infinite case in the regular Cesaro mean theorem either).

Any help would be greatly appreciated.

P.S ~ I'm loving the chance to finally use $\LaTeX$ :)

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    $\begingroup$ If $t_n = 1$ then the usual Cesaro mean is recovered, so perhaps a proof along these lines can be mimicked? (try replacing $n$ in the linked proof with $\sum_1^n t_k$) $\endgroup$ – Antonio Vargas Nov 16 '12 at 17:18
  • $\begingroup$ I've already looked at this proof, but did not really understand the notation. The idea appears to be similar to my own proof however. I replaced n with that $t_n$ sum but I'm having trouble controlling the $t_n a_n$ sum.. $\endgroup$ – Adar Hefer Nov 16 '12 at 17:27

If $L\in \mathbb{R}$ this result holds. Fix $\varepsilon>0$. Since $$ \lim\limits_{k\to\infty}a_k=L $$ then there exist $N\in\mathbb{N}$ such that $|a_k-L|<\varepsilon/2$ for all $k>K$. Since $$ \lim\limits_{n\to\infty}\sum\limits_{k=1}^n t_k=+\infty $$ then there exist $N\in\mathbb{N}$ such that $$ \sum\limits_{k=1}^n t_k>2\varepsilon^{-1}\left|\sum\limits_{k=1}^K t_k(a_k-L)\right| $$ for all $n>N$. In this case for all $n>\max(N,K)$ we have $$ \begin{align} |b_n-L| &=\left|\left(\sum\limits_{k=1}^n t_k\right)^{-1}\sum\limits_{k=1}^n a_k t_k-L\right|\\ &=\left(\sum\limits_{k=1}^n t_k\right)^{-1}\left|\sum\limits_{k=1}^n a_k t_k-L\sum\limits_{k=1}^n t_k\right|\\ &=\left(\sum\limits_{k=1}^n t_k\right)^{-1}\left|\sum\limits_{k=1}^n t_k(a_k-L)\right|\\ &=\left(\sum\limits_{k=1}^n t_k\right)^{-1}\left|\sum\limits_{k=1}^K t_k(a_k-L)+\sum\limits_{k=K+1}^n t_k(a_k-L)\right|\\ &\leq\left(\sum\limits_{k=1}^n t_k\right)^{-1}\left(\left|\sum\limits_{k=1}^K t_k(a_k-L)\right|+\left|\sum\limits_{k=K+1}^n t_k(a_k-L)\right|\right)\\ &\leq\left(\sum\limits_{k=1}^n t_k\right)^{-1}\left|\sum\limits_{k=1}^K t_k(a_k-L)\right|+\left(\sum\limits_{k=1}^n t_k\right)^{-1}\sum\limits_{k=K+1}^n t_k|a_k-L|\\ &\leq\varepsilon/2+\left(\sum\limits_{k=1}^n t_k\right)^{-1}\sum\limits_{k=K+1}^n t_k\varepsilon/2\\ &\leq\varepsilon/2+\varepsilon/2=\varepsilon \end{align} $$ Since $\varepsilon>0$ is arbitrary we showed that $$ \lim\limits_{n\to\infty} b_n=L $$

If $L=\infty$ there is a counterexample. Consider case $t_k=k$ and $a_k=(-1)^k k$, then $$ \lim\limits_{k\to\infty} a_k=\infty\\ \lim\limits_{n\to\infty}\sum\limits_{k=1}^n t_k=\lim\limits_{n\to\infty}\frac{n(n+1)}{2}=\infty $$ Moreover $$ \sum\limits_{k=1}^n t_ka_k=\sum\limits_{k=1}^n (-1)^k k^2=\frac{(-1)^n}{2}n(n+1)\\ b_n=\frac{\sum\limits_{k=1}^n t_ka_k}{\sum\limits_{k=1}^n t_k}= \frac{(-1)^n}{2} $$ So $\lim\limits_{n\to\infty} b_n$ even doesn't exist, not to mention it is equals to $L$.

  • $\begingroup$ Awesome. I feel a little dumb for missing one of the transitions in the middle. Thank you! You wouldn't happen to know the version for $a_n \to \infty$? $\endgroup$ – Adar Hefer Nov 16 '12 at 18:41
  • $\begingroup$ @AdarHefer See edits, to my answer. $\endgroup$ – Norbert Nov 17 '12 at 10:19
  • $\begingroup$ I thought (-1)^k * k does not exist..? the exercise specifically says it holds for the infinite case as well. I don't think a counter-example exists. $\endgroup$ – Adar Hefer Nov 17 '12 at 20:48
  • $\begingroup$ $\lim\limits_{k\to\infty} (-1)^k k=\infty$, while $\lim\limits_{k\to\infty} k=+\infty$. So my question is equality $\lim\limits_{k\to\infty}a_k=\infty$ should be understood as $\lim\limits_{k\to\infty}a_k=+\infty$? $\endgroup$ – Norbert Nov 17 '12 at 23:17
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    $\begingroup$ I have three definitions of infinite limits ($=\infty$,$=-\infty$,$=+\infty$) $\endgroup$ – Norbert Dec 1 '12 at 12:18

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