calculus - given Sn, find a_n and sum of a_n

I am given $$s_n=\frac{n-1}{n+1}$$ find $a_n$ and $\sum_{n=1}^\infty a_n$

I use $a_n = s_n-s_{n-1}$ and got $$lim_{n->{\infty}}\frac{2}{n(n+1)}$$

Then the theorem says

$$\sum_{n=1}^\infty a_n= lim_{n->{\infty}}s_n$$

and the answer is 1 as n approaches infinity.

I also know that if I change $a_n$ into a telescope sum and solve it.

$$\lim_{n->\infty} \frac{2}{n}-\frac{2}{n+1}$$

it becomes $$\lim_{n->\infty}2-\frac{2}{n+1}$$

the answer is 2 as n approach infinity.

now, I don't understand why the two are not identical. did I make theoretical error with my reasoning?