Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+2n+2}$ is convergent.
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{n^2+2n+2}\]
\[\Rightarrow a_n=\frac{1}{n^2+2n+2}=\frac{1}{(n+1)^2+1}\]
$a_n$ is positive term and monotonically decreasing.
So Integral test is applicable.
Let \[I=\int_{1}^{\infty}\frac{1}{(x+1)^2+1}dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{(x+1)^2+1}dx\;\;\;\;\;\;\;\;\;\;\ldots(1)\]
Let \[I_1=\int\frac{1}{(x+1)^2+1}dx\]
Substitute $y=x+1\;\Rightarrow\; dy=dx$
\[\Rightarrow I_1=\int\frac{dy}{y^2+1}\]
\[\left[\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\right]\]
\[\Rightarrow I_1=\tan^{-1}y\]
\[\Rightarrow I_1=\tan^{-1}(x+1)\;\;\;\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow\infty}\left[\tan^{-1}(x+1)\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[\tan^{-1}(t+1)-\tan^{-1}2\right]\]
\[\Rightarrow I=\frac{\pi}{2}-\tan^{-1}2\]
Which is finite so $I$ is convergent.
By integral test, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+2n+2}$ is convergent.
