Question

The value of the expression $\tan (\tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{2}{9}) + \tan^{-1}(\frac{1}{8}) + \tan^{-1}(\frac{2}{25}) + \tan^{-1}(\frac{1}{18}) + ......\infty)$, is:

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Answer 2

We note that

$$\tan^{-1}(n+1)-\tan^{-1}(n-1)=\tan^{-1}\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)=\tan^{-1}\left(\frac{2}{n^2}\right).$$

Thus, taking $n=2,3,4,\ldots$, we have:

$$\tan^{-1}\left(\frac{1}{2}\right)=\tan^{-1}(3)-\tan^{-1}(1),\quad \tan^{-1}\left(\frac{2}{9}\right)=\tan^{-1}(4)-\tan^{-1}(2),\quad \text{etc.}$$

So the given sum becomes a telescoping series:

$$S=\sum_{n=2}^{\infty}\left[\tan^{-1}(n+1)-\tan^{-1}(n-1)\right].$$

Writing out the terms:

$$S=(\tan^{-1}(3)-\tan^{-1}(1)) + (\tan^{-1}(4)-\tan^{-1}(2)) + (\tan^{-1}(5)-\tan^{-1}(3)) + \cdots.$$

Most terms cancel, leaving:

$$S=\lim_{N\to\infty}\Big[\tan^{-1}(N+1)+\tan^{-1}(N)-\tan^{-1}(1)-\tan^{-1}(2)\Big].$$

Since $\tan^{-1}(N)\to \pi/2$ as $N\to \infty$, we have:

$$S=\pi -\left(\tan^{-1}(1)+\tan^{-1}(2)\right).$$

But $\tan^{-1}(1)=\frac{\pi}{4}$, hence:

$$S=\pi-\left(\frac{\pi}{4}+\tan^{-1}(2)\right)=\frac{3\pi}{4}-\tan^{-1}(2).$$

We now find:

$$\tan S=\tan\Bigl(\frac{3\pi}{4}-\tan^{-1}(2)\Bigr).$$

Using the formula $\tan(A - B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B}$, with $A=\frac{3\pi}{4}$ (where $\tan\frac{3\pi}{4}=-1$) and $B=\tan^{-1}(2)$ (so that $\tan B=2$):

$$\tan S=\frac{-1-2}{1+(-1)(2)}=\frac{-3}{1-2}=\frac{-3}{-1}=3.$$