\({ }^{\infty} \sum_{r=1} \tan ^{-1}\left(1 / 2 r^{2}\right)=\) (a) \((\pi / 4)\) (b) \((\pi / 2)\) (c) \(\tan ^{-1}(\mathrm{n})-(\pi / 4)\) (d) \(\tan ^{-1}(n+1)-(\pi / 4)\)

First, we will write down the first few terms of the sum to check for a pattern. The given sum is: \[\sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2 r^{2}}\right)\] The first few terms of the sum are: \(r = 1\) -> \(\tan^{-1}\left(\frac{1}{2}\right)\) \(r = 2\) -> \(\tan^{-1}\left(\frac{1}{8}\right)\) \(r = 3\) -> \(\tan^{-1}\left(\frac{1}{18}\right)\)

Upon inspection, we notice that the given sum does not have an obvious pattern or closed-form formula. Therefore, we will need to compare our sum with the given options and try to evaluate the first few terms of each option to match with the given sum. Option (a): \(\frac{\pi}{4}\) This option is a constant and does not resemble our sum. Option (b): \(\frac{\pi}{2}\) This option is also a constant and does not resemble our sum. Option (c): \(\tan^{-1}(n) - \frac{\pi}{4}\) Let's check the first term by substituting \(n = 1\): \[\tan^{-1}(1) - \frac{\pi}{4} = \frac{\pi}{4} - \frac{\pi}{4} = 0\] This option evaluates to zero for the first term and does not match our sum. Option (d): \(\tan^{-1}(n+1) - \frac{\pi}{4}\) Let's check the first term by substituting \(n = 1\): \[\tan^{-1}(2) - \frac{\pi}{4}\] Now, we know \(\tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4}\) (as seen in the first term of our sum), and also : \[\tan(2) = \frac{2\tan\left(\frac{1}{2}\right)}{1-\tan^{2}\left(\frac{1}{2}\right)}\] So, \(\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}(2) = \frac{\pi}{2}\) Thus, \(\tan^{-1}(2) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\) Therefore, the first term in option (d) is \(\tan^{-1}(2) - \frac{\pi}{4} = \tan^{-1}\left(\frac{1}{2}\right)\), which matches the first term of the given sum. By induction, we can show that this pattern holds for all the sum terms: Consider the \(k\_th\) term of option (d): \[\tan^{-1}(k+1) - \frac{\pi}{4}\] Using the same logic applied for the first term, we can show that: \[\tan^{-1}\left(\frac{1}{2k^{2}}\right) = \tan^{-1}(k+1) - \frac{\pi}{4}\] Hence, option (d) matches the pattern of the given sum, and the sum can be written as: \[\sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2 r^{2}}\right) = \tan^{-1}(n+1) - \frac{\pi}{4}\] The correct answer is (d) \(\tan^{-1}(n+1)-\frac{\pi}{4}\).