Question:
If
\[
\sum_{r=1}^{\infty}\tan^{-1}\left(\frac{1}{2r^{2}}\right)=a,
\]
then \(\tan a\) is equal to:
If
\[
\sum_{r=1}^{\infty}\tan^{-1}\left(\frac{1}{2r^{2}}\right)=a,
\]
then \(\tan a\) is equal to:
Show Hint
The argument factorization trick $\frac{1}{2r^2} \rightarrow \frac{2}{1+4r^2-1}$ is a standard method used to solve inverse tangent series in engineering math exams. Recognizing this step helps you quickly convert the sum into a clean telescoping form.
Updated On: May 28, 2026
- $1$
- $0$
- $\sqrt{3}$
- $\frac{\pi}{4}$
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Infinite series involving inverse trigonometric functions are typically solved by transforming the general term into a telescoping difference format using the standard identity:
$$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$$
Step 1: Deconstruct the general term of the summation series.
The general term for any index $r$ is given by $T_r = \tan^{-1}\left(\frac{1}{2r^2}\right)$. Let us multiply the numerator and denominator parameters inside the argument by $2$: $$T_r = \tan^{-1}\left(\frac{2}{4r^2}\right)$$ Add and subtract $1$ inside the denominator to match the structure of our inverse tangent difference identity: $$T_r = \tan^{-1}\left(\frac{2}{1 + (4r^2 - 1)}\right) = \tan^{-1}\left(\frac{2}{1 + (2r - 1)(2r + 1)}\right)$$
Step 2: Express as a telescoping difference layout.
Notice that the difference between the two factors in the denominator matches the numerator exactly: $(2r + 1) - (2r - 1) = 2$. We can rewrite $T_r$ as: $$T_r = \tan^{-1}\left(\frac{(2r + 1) - (2r - 1)}{1 + (2r - 1)(2r + 1)}\right) = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1)$$
Step 3: Sum the terms to find the partial summation $S_n$.
Let us write out the expansion for the partial sum $S_n = \sum_{r=1}^n T_r$: $$S_n = \left[\tan^{-1}(3) - \tan^{-1}(1)\right] + \left[\tan^{-1}(5) - \tan^{-1}(3)\right] + \dots + \left[\tan^{-1}(2n+1) - \tan^{-1}(2n-1)\right]$$ Cancelling out matching interior terms leaves only the first and last elements: $$S_n = \tan^{-1}(2n + 1) - \tan^{-1}(1) = \tan^{-1}(2n + 1) - \frac{\pi}{4}$$
Step 4: Evaluate the infinite limit sum to calculate $\tan a$.
To find the total infinite sum value $a$, evaluate the limit as $n \rightarrow \infty$: $$a = \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left[ \tan^{-1}(2n + 1) - \frac{\pi}{4} \right] = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$ Now, evaluate the requested functional value $\tan a$: $$\tan a = \tan\left(\frac{\pi}{4}\right) = 1$$ This matches option (A) perfectly.
The general term for any index $r$ is given by $T_r = \tan^{-1}\left(\frac{1}{2r^2}\right)$. Let us multiply the numerator and denominator parameters inside the argument by $2$: $$T_r = \tan^{-1}\left(\frac{2}{4r^2}\right)$$ Add and subtract $1$ inside the denominator to match the structure of our inverse tangent difference identity: $$T_r = \tan^{-1}\left(\frac{2}{1 + (4r^2 - 1)}\right) = \tan^{-1}\left(\frac{2}{1 + (2r - 1)(2r + 1)}\right)$$
Step 2: Express as a telescoping difference layout.
Notice that the difference between the two factors in the denominator matches the numerator exactly: $(2r + 1) - (2r - 1) = 2$. We can rewrite $T_r$ as: $$T_r = \tan^{-1}\left(\frac{(2r + 1) - (2r - 1)}{1 + (2r - 1)(2r + 1)}\right) = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1)$$
Step 3: Sum the terms to find the partial summation $S_n$.
Let us write out the expansion for the partial sum $S_n = \sum_{r=1}^n T_r$: $$S_n = \left[\tan^{-1}(3) - \tan^{-1}(1)\right] + \left[\tan^{-1}(5) - \tan^{-1}(3)\right] + \dots + \left[\tan^{-1}(2n+1) - \tan^{-1}(2n-1)\right]$$ Cancelling out matching interior terms leaves only the first and last elements: $$S_n = \tan^{-1}(2n + 1) - \tan^{-1}(1) = \tan^{-1}(2n + 1) - \frac{\pi}{4}$$
Step 4: Evaluate the infinite limit sum to calculate $\tan a$.
To find the total infinite sum value $a$, evaluate the limit as $n \rightarrow \infty$: $$a = \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left[ \tan^{-1}(2n + 1) - \frac{\pi}{4} \right] = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$ Now, evaluate the requested functional value $\tan a$: $$\tan a = \tan\left(\frac{\pi}{4}\right) = 1$$ This matches option (A) perfectly.
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