calculus - General Cesaro summation with weight

Assume that $a_n\to \ell $ is a convergent sequence of complex numbers and $\{\lambda_n\}$ is a sequence of positive real numbers such that $\sum\limits_{k=0}^{\infty}\lambda_k = \infty$

Then, show that,
$$\lim_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k} \sum_\limits{k=0}^{n}\lambda_k a_k=\ell =\lim_{n\to\infty} a_n$$

(Note that : This is more general than the special case where, $\lambda_n= 1$)

Answer

Let $\varepsilon >0$ and $N$such that $|a_k-l|\le \varepsilon $ for all $k>N$
Then, for $n>N$ we have,

\begin{split}\left| \frac{\sum_\limits{k=0}^{n}\lambda_k a_k}{\sum_\limits{k=0}^{n}\lambda_k} -l\right|
&= &\left| \frac{\sum_\limits{k=0}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\
&= &\left| \frac{\sum_\limits{k=0}^{N}\lambda_k (a_k - l)+\sum_\limits{k=N}^{n}\lambda_k (a_k - l)}{\sum_\limits{k=0}^{n}\lambda_k} \right|\\
&\le & \frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \frac{\sum_\limits{k=N}^{n}\lambda_k \underbrace{\left| a_k - l\right|}_{\le\varepsilon}}{\sum_\limits{k=0}^{n}\lambda_k} \\
&\le&
\frac{M}{\sum_\limits{k=0}^{n}\lambda_k} + \varepsilon\to 0
\end{split}
since $\sum_\limits{k=0}^{N}\lambda_k\to \infty$.
Where $M= \left|\sum_\limits{k=0}^{N}\lambda_k( a_k-l)\right|$