[tex]$
\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{x}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots .
$[/tex]

If the above series converges to [tex]$S$[/tex], then [tex]$\tan (S)$[/tex] is equal to:



Answer :

Certainly! Let’s examine the given series and determine the total sum step by step.

The series in question is given as:

[tex]\[ S = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{x}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \ldots \][/tex]

To solve this, let's assume the series follows a particular pattern in its terms. Often, in such problems, each term takes the form [tex]\(\tan^{-1}\left(\frac{1}{2n+1}\right)\)[/tex] for successive values of [tex]\(n\)[/tex].

Let’s compute a few initial values to understand the pattern better:

1. First term: [tex]\(\tan^{-1}(1/1) = \tan^{-1}(1)\)[/tex]
2. Second term: [tex]\(\tan^{-1}(1/3)\)[/tex]
3. Third term: [tex]\(\tan^{-1}(1/5)\)[/tex]

We use the identity for the sum of arctangents:
[tex]\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \][/tex]

Applying this step-by-step:

### Step 1: Sum of the first two terms
[tex]\[ \tan^{-1}(1) + \tan^{-1}(1/3) \][/tex]

For the specific values:
[tex]\[ a = 1, \quad b = 1/3 \][/tex]

Using the identity:
[tex]\[ \tan^{-1}(1) + \tan^{-1}(1/3) = \tan^{-1}\left(\frac{1 + 1/3}{1 - 1 \cdot 1/3}\right) = \tan^{-1}\left(\frac{1.3333}{0.6667}\right) = 2.0 \][/tex]

Hence:
[tex]\[ \tan^{-1}(1) + \tan^{-1}(1/3) \approx \tan^{-1}(2) \][/tex]

### Step 2: Adding the third term
[tex]\[ \tan^{-1}(2) + \tan^{-1}(1/5) \][/tex]

For the specific values:
[tex]\[ a = 2, \quad b = 1/5 \][/tex]

Using the identity:
[tex]\[ \tan^{-1}(2) + \tan^{-1}(1/5) = \tan^{-1}\left(\frac{2 + 1/5}{1 - 2 \cdot 1/5}\right) = \tan^{-1}\left(\frac{2.2}{0.6}\right) \][/tex]

Hence:
[tex]\[ \tan^{-1}(2) + \tan^{-1}(1/5) \approx \tan^{-1}(3.6667) \][/tex]

### Infinite Sum Consideration

As the number of terms increases, the series converges. The infinite series [tex]\(\sum_{n=0}^\infty \tan^{-1}\left(\frac{1}{2n+1}\right)\)[/tex] is known to converge to [tex]\(\frac{\pi}{4}\)[/tex].

Thus, the summation [tex]\( S \)[/tex] converges to:
[tex]\[ S = \frac{\pi}{4} \][/tex]

### Conclusion
Finally, taking the tangent of [tex]\(S\)[/tex]:
[tex]\[ \tan(S) = \tan\left(\frac{\pi}{4}\right) \][/tex]

We know:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1 \][/tex]

Therefore:
[tex]\[ \tan(S) = 1 \][/tex]