Answer
$b_n$ converges to $\frac{\pi}{4}$.
Work Step by Step
We have
$$
\lim _{n \rightarrow \infty} b_n=\lim _{n \rightarrow \infty}
\tan ^{-1}\left(\frac{n+2}{n+5}\right)
\\
=\lim _{n \rightarrow \infty}
\tan ^{-1}\left(\frac{1+(2/n)}{1+(5/n)}\right)=\tan ^{-1}\left(1\right)=\frac{\pi}{4}.
$$
So, $b_n$ converges to $\frac{\pi}{4}$.
