I have to calculate the series \sum_{k=2}^\infty(\frac{1}{k}-\frac{1}{k+2}) Using the definition: L=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^n a^k Obviously \lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n+2})=0,

As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
${\displaystyle \sum _2^{\mathrm{\infty }}(\frac{1}{k}-\frac{1}{k+2})=(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{6}-\frac{1}{8})+\dots }$
From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the ${\displaystyle \frac{1}{2}}$ and ${\displaystyle \frac{1}{3}}$.
Therefore, the answer is ${\displaystyle \frac{5}{6}}$.

Another, Different way that uses limit like you wanted to is to find the partial sum! You first compute, $S_2=\frac{1}{2}-\frac{1}{4},S_3=S_2+(\frac{1}{3}-\frac{1}{5})$ and so on... What you will find (and you can prove this fact by induction) is that
$S_n=\frac{5n^2+3n-8}{6(n+1)(n+2)}$
Now,
$\underset{n\to \mathrm{\infty }}{lim}\frac{5n^2+3n-8}{6n^2+18n+12}=\frac{5}{6}$