question: find out if the following converges and if it does find what number it converges to: $\displaystyle \sum_{n=1}^{\infty} \frac{2}{n^2+4n+3}$ use a telescoping sum

work: I used partial fractions to get $\displaystyle \frac{2}{n^2+4n+3}=\frac{1}{n+1}-\frac{1}{n+3}$

then I used a telescoping sum and the first term and last term remained, which are $\displaystyle \frac{1}{2}+-\frac{1}{n+3}$ and then taking the limit n-->infinity I'm left with 1/2.

Is this right? When I graphed the function it appears to go to 1.