Show that a sequence $\{s_n\}$ converges to a limit $L$ if and only if the sequence $\{s_n-L\}$ converges to zero. I'm looking for hints and ideas on how to improve the draft of my proof.

Proof: Let $\{s_n\}$ be any given sequence. Then take the limit of the sequence {$s_n$} as shown.

$$\lim_{n\to \infty} s_n=L $$

Indeed, as $n$ tends to infinity, the sequence converges to $L$.

Once more, we take the difference of the sequence $\{s_n\}$ and the limit $L$.

Lastly, we take the limit of $\{s_n -L\}$, so 

$$\lim_{n\to \infty}s_n - L = 0$$

Therefore, the sequence $\{s_n - L\}$ converges to zero.
Hint: First of all, you need a clear definition of what it means for a sequence of real numbers to converge to a limit.

Secondly, using this definition of convergence, show that $s_n$ converges to $L$, if one assumes that $s_n{}-{}L$ converges to $0$.

Finally, using this definition of convergence again, show that $s_n{}-{}L$ converges to $0$, if one assumes that $s_n$ converges to $L$.

This follows the typical structure of an if, and only if, proof Wikipedia.
Let $\epsilon &gt; 0 $ be given. By definition, $s_n \to L$ means that  a positive integer $N$ can be chosen so that for all $n &gt; N $, we have 

$$ |s_n - L | &lt; \epsilon $$

Let $a_n = s_n - L $. Hence, $a_n \to 0 $ or $s_n - L \to 0 $. Similarly, you prove the other direction.