A generalized Cesaro Mean.. any ideas? 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 $\LaTeX$   :)
 A: 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 $L$.
