Show that $\sum_{k=1}^\infty (a_k - a_{k+1}) = a_1 - l$ if $\lim_{k \to \infty} a_k = l$ 
Show that $$\sum_{k=1}^\infty (a_k - a_{k+1}) = a_1 - l$$ if $$\lim_{k \to \infty} a_k = l$$

Can anyone please provide me with hints as to go about solving this problem?
 A: $(a_1 - a_2) + (a_2 - a_3) + \ldots + (a_N - a_{N+1}) = a_1 - a_{N+1}$.
Just do $N \rightarrow \infty$.
A: Hint: This is a telescoping series
$$
\begin{align}
\sum_{k=1}^n(a_k-a_{k+1})
&=\sum_{k=1}^na_k-\sum_{k=1}^na_{k+1}\\
&=\sum_{k=1}^na_k-\sum_{k=2}^{n+1}a_k\\
&=\left(a_1+\sum_{k=2}^na_k\right)-\left(a_{n+1}+\sum_{k=2}^na_k\right)\\[6pt]
&=a_1-a_{n+1}
\end{align}
$$
A: Considering the partial sums of the series:
$$
\begin{align}
\sum_{k=1}^n(a_k-a_{k+1})
&=\sum_{k=1}^na_k-\sum_{k=2}^{n+1}a_k\\
&=a_1+\sum_{k=2}^na_k-\sum_{k=2}^{n}a_k-a_{n+1}\\
&=a_1-a_{n+1}
\end{align}
$$
Now  the sequence $(a_{n+1})_{n \in \mathbb{N}}$ is a sub-sequence of $(a_{n})_{n \in \mathbb{N}}$ and so shares the same same limit.
Finally, by the algebra of limits, we have,
$$
\begin{align}
\lim_{n \rightarrow \infty} \left( \sum_{k=1}^n(a_k-a_{k+1}) \right) 
&= \lim_{n \rightarrow \infty} \left(a_1-a_{n+1}\right)\\
&= \lim_{n \rightarrow \infty} (a_1) - \lim_{n \rightarrow \infty} (a_{n+1})\\
&= a_1 - l\\
\end{align}
$$
