Infinite Series Value Radians
What is the value of $ \sum_{n=1}^\infty (\tan^{-1}\sqrt{n}-\tan^{-1}\sqrt{n+1}) $?
Your answer should be in radians.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
The $ k $th partial sum is
\begin{align*}
&\sum_{n = 1}^k (\tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n + 1}) \\
&= (\tan^{-1} 1 - \tan^{-1} \sqrt{2}) + (\tan^{-1} \sqrt{2} - \tan^{-1} \sqrt{3}) \\
&\quad + (\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{4}) + \dots + (\tan^{-1} \sqrt{k} - \tan^{-1} \sqrt{k + 1}) \\
&= \tan^{-1} 1 - \tan^{-1} \sqrt{k + 1} \\
&= \frac{\pi}{4} - \tan^{-1} \sqrt{k + 1}.\end{align*}
As $ k $ goes to infinity, $ \tan^{-1} \sqrt{k + 1} $ approaches $ \frac{\pi}{2}, $ so the limit of the sum as $ n $ goes to infinity is $ \frac{\pi}{4} - \frac{\pi}{2} = \boxed{-\frac{\pi}{4}} $.
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