## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 15

#### 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 $\frac{\pi}{4}$.

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