limit of arithmetic weighted mean ($\lim_\limits{n\to\infty}\frac{\sum_\limits{k=1}^n{t_ka_n}}{\sum_\limits{k=1}^n{t_k}} = L$) Given that:
$t_n>0 \\ \lim_\limits{n\to\infty}\sum_\limits{k=1}^n{t_k} = \infty$
$\lim_\limits{n\to\infty}a_n = L$
(either $L=\pm\infty$ or $L\in\mathbb{R}$)
I'm trying to prove that:
$\lim_\limits{n\to\infty}\frac{\sum_\limits{k=1}^n{t_ka_n}}{\sum_\limits{k=1}^n{t_k}} = L$
Of course, it seems very similar Cesaro mean, but I've had trouble trying to apply a similar method here.
Any idea? Thanks!
 A: Assume $a_n\to L$.
If $t_k>0$ and $\sum_{1}^n t_k\to \infty$, then:
$$\frac{A}{\sum_1^n t_k}\to 0$$ for any constant $A$.
From this we see that, for any $N>0$, letting:
$$r_{N,n}=\frac{\sum_{k=N+1}^{N+n} t_k}{\sum_{k=0}^{N+n} t_k}$$
then $\lim_{n\to\infty} r_{N,n} = 1$, since $r_{N,n}=1-\frac{A}{\sum_{1}^{N+n} t_k}$ where $A=\sum_{1}^N t_k$ is constant.
Pick $N>0$ so that $|a_n-L|<\epsilon/3$ for $n>N$. Pick $N'>N$ so that when $n>N'$, $L|r_n-1|<\epsilon/3$. Finally, set $A_N=\sum_{k=1}^{N} t_ka_k$, and find $N''>N'$ so that if $n>N''$, then $\left|\frac{A_N}{\sum_1^n t_k}\right|<\epsilon/3$.
We have:
$$\sum_{k=1}^{n}t_ka_k = \sum_{k=1}^{N} t_ka_k + \sum_{k=N+1}^{n} t_k(a_k-L) + L\sum_{k=N+1}^{n}t_k$$
Then, for $n>N''$:
$$\frac{\sum_{1}^{n}t_ka_k}{\sum_1^n t_k}-L = \frac{A_N}{\sum_{1}^n t_k} + \frac{\sum_{N+1}^{n} t_k(a_k-L)}{\sum_{1}^n t_k}+L(r_{N,n}-1)$$
The we get that:
$$\left|\frac{\sum_{1}^{n}t_ka_k}{\sum_1^n t_k}-L\right|<\epsilon$$
The case of $L=\pm\infty$ has to be handled separately. 
In the case $L=+\infty$, let $M>0$ and find $N$ such that $a_n>3M$ for $n>N$.
Then find $N'>N$ such that $r_n>\frac{2}{3}$ and find $N''>N'$ such that for $n>N''$, $\left|\frac{A_N}{\sum_1^n t_k}\right|<M$.
Then you get, for $n>N''$:
$$\frac{\sum_{1}^{n} t_ka_k}{\sum_1^n t_k} > -M+ \frac{2}{3}\cdot 3M=M$$
A: Hint. For the case where $L$ is finite.
Pick-up $N$ such that for $n >N$ you have
$$\begin{aligned}
\left\vert \frac{\sum_{k=1}^n{t_ka_n}}{\sum_{k=1}^n{t_k}} -L \right\vert &\le \frac{\sum_{k=1}^n{t_k \vert a_n-L \vert}}{\sum_{k=1}^n{t_k}}\\
&\le \frac{\sum_{k=1}^N{t_k \vert a_n-L \vert}}{\sum_{k=1}^n{t_k}} + \frac{\sum_{k=N+1}^n{t_k \vert a_n-L \vert}}{\sum_{k=1}^n{t_k}}\\
&\le \frac{\sum_{k=1}^N{t_k \vert a_n-L \vert}}{\sum_{k=1}^n{t_k}} + \frac{\epsilon}{2}
\end{aligned}$$
As $\lim_{n\to\infty}\sum_{k=1}^n{t_k} = \infty$, you can find $M > N$ such that for $n > M$
$$\frac{\sum_{k=1}^N{t_k \vert a_n-L \vert}}{\sum_{k=1}^n{t_k}} \le \frac{\epsilon}{2}$$
For $n>m>N$ you have therefore
$$\left\vert \frac{\sum_{k=1}^n{t_ka_n}}{\sum_{k=1}^n{t_k}} -L \right\vert \le \epsilon$$
If $L = +\infty$, by a similar method, prove that $\frac{\sum_{k=1}^n{t_ka_n}}{\sum_{k=1}^n{t_k}}$ can be as large as desired.
