Determine if the series converges or diverges. If the series converges, find its sum.
-$$\sum _ { n = 1 } ^ { \infty } \left( \tan ^ { - 1 } ( n + 1 ) - \tan ^ { - 1 } n \right)$$
A) converges; $\frac { \pi } { 2 }$
B) converges; $\pi$
C) diverges
D) converges; $- \frac { \pi } { 2 }$

Question

Quizplus · Accepted Answer

We can use the identity $	an^{-1} (n+1) - 	an^{-1} n = 	an^{-1} \left(\frac{n+1}{1+n(n+1)}ight)$. Note that $\frac{n+1}{1+n(n+1)}$ is decreasing as $n$ increases, so we can apply the integral test. $$\int_1^\infty 	an^{-1} \left(\frac{n+1}{1+n(n+1)}ight) dn = \int_2^\infty 	an^{-1} \left(\frac{n}{1+n(n-1)}ight) dn$$ We can now use the substitution $u = \frac{1}{n}$ to get $$\int_0^1 \frac{	an^{-1}(1/u)}{u^2+2u(1-u)} du = \int_0^1 \frac{	an^{-1} u}{u(1-u)} du.$$ We can use partial fractions to write this as $$\int_0^1 \left( \frac{1}{u} - \frac{1}{1-u} ight) \frac{	an^{-1}u}{du}.$$ Using the fact that $\int_0^1 \frac{1}{u(1-u)} du = \infty$, we can conclude that the given series converges. To find its sum, we evaluate the integral as follows: $$\int_0^1 \left( \frac{1}{u} - \frac{1}{1-u} ight) \frac{	an^{-1}u}{du} = \int_0^1 \frac{	an^{-1} u}{u} du - \int_0^1 \frac{	an^{-1} u}{1-u} du$$$$= \frac{\pi}{2} \ln 2 + \frac{\pi^2}{16}.$$ Therefore, the answer is A.

Determine If the Series Converges or Diverges ∑n=1∞(tan⁡−1(n+1)−tan⁡−1n)\sum _ { n = 1 } ^ { \infty } \left( \tan ^ { - 1 } ( n + 1 ) - \tan ^ { - 1 } n \right)n=1∑∞​(tan−1(n+1)−tan−1n)

Determine If the Series Converges or Diverges $\sum _ { n = 1 } ^ { \infty } \left( \tan ^ { - 1 } ( n + 1 ) - \tan ^ { - 1 } n \right)$