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Question: $\sum_{r=0}^{\infty} \tan^{-1} \left( \frac{r((r+1)!)}{(r+1)+((r+1)!)^2} \right)$ is equal to:...

∑r=0∞tan⁡−1(r((r+1)!)(r+1)+((r+1)!)2)\sum_{r=0}^{\infty} \tan^{-1} \left( \frac{r((r+1)!)}{(r+1)+((r+1)!)^2} \right) is equal to:

A

π4\frac{\pi}{4}

Answer

π4\frac{\pi}{4}

Explanation

Solution

Let the given sum be SS. The general term of the sum is ar=tan⁡−1(r((r+1)!)(r+1)+((r+1)!)2)a_r = \tan^{-1} \left( \frac{r((r+1)!)}{(r+1)+((r+1)!)^2} \right). We want to express the argument of tan⁡−1\tan^{-1} in the form x−y1+xy\frac{x-y}{1+xy} to use the identity tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right).

The argument is r((r+1)!)(r+1)+((r+1)!)2\frac{r((r+1)!)}{(r+1)+((r+1)!)^2}.

After significant manipulation (omitted for brevity), we can rewrite the general term as:

ar=tan⁡−1(1r!)−tan⁡−1(1(r+1)!)a_r = \tan^{-1} \left( \frac{1}{r!} \right) - \tan^{-1} \left( \frac{1}{(r+1)!} \right)

The sum is S=∑r=0∞ar=∑r=0∞(tan⁡−1(1r!)−tan⁡−1(1(r+1)!))S = \sum_{r=0}^{\infty} a_r = \sum_{r=0}^{\infty} \left( \tan^{-1} \left( \frac{1}{r!} \right) - \tan^{-1} \left( \frac{1}{(r+1)!} \right) \right). This is a telescoping series.

The NN-th partial sum is SN=∑r=0N(tan⁡−1(1r!)−tan⁡−1(1(r+1)!))S_N = \sum_{r=0}^{N} \left( \tan^{-1} \left( \frac{1}{r!} \right) - \tan^{-1} \left( \frac{1}{(r+1)!} \right) \right). SN=(tan⁡−1(10!)−tan⁡−1(11!))+(tan⁡−1(11!)−tan⁡−1(12!))+⋯+(tan⁡−1(1N!)−tan⁡−1(1(N+1)!))S_N = \left(\tan^{-1}\left(\frac{1}{0!}\right) - \tan^{-1}\left(\frac{1}{1!}\right)\right) + \left(\tan^{-1}\left(\frac{1}{1!}\right) - \tan^{-1}\left(\frac{1}{2!}\right)\right) + \dots + \left(\tan^{-1}\left(\frac{1}{N!}\right) - \tan^{-1}\left(\frac{1}{(N+1)!}\right)\right). SN=tan⁡−1(10!)−tan⁡−1(1(N+1)!)=tan⁡−1(1)−tan⁡−1(1(N+1)!)S_N = \tan^{-1}\left(\frac{1}{0!}\right) - \tan^{-1}\left(\frac{1}{(N+1)!}\right) = \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{(N+1)!}\right). As N→∞N \to \infty, 1(N+1)!→0\frac{1}{(N+1)!} \to 0. S=lim⁡N→∞SN=tan⁡−1(1)−tan⁡−1(0)=π4−0=π4S = \lim_{N \to \infty} S_N = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}.