The sum of the infinite series cot⁻¹(7/4) + cot⁻¹(19/4) + cot⁻¹(39/4) + cot⁻¹(67/4) + ... is :- (1)

Q: Why did we assume the numerator follows a quadratic pattern? Could it be something else?

In competitive exams like JEE, when you see a sequence where the second difference is constant, it is mathematically guaranteed to be a quadratic. It is the most elegant way to define the progression of terms!

Q: I got stuck at the denominator. How do I know to add and subtract 1?

That is the 'telescoping' secret! We always look for the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)$. By forcing a '$1$' into the denominator, we unlock the ability to split the term into a difference of two inverse tangents.

Q: Why does the sum telescope? I keep getting confused about which terms cancel.

Think of it like a line of dominoes. When you write $T_n = f(n) - f(n-1)$, the $f(n-1)$ of the second term cancels the $f(n-1)$ of the first. Only the very first negative term and the very last positive term remain standing. It is a beautiful, clean collapse!

Q: What happens to the $\tan^{-1}(n + 1/2)$ as $n$ goes to infinity?

As $n$ grows without bound, the argument $(n + 1/2)$ approaches infinity. Since the graph of $\tan^{-1}(x)$ has a horizontal asymptote at $\pi/2$, the value of the function approaches $\pi/2$.

Q: Is $\pi/2 - \tan^{-1}(1/2)$ the same as $\cot^{-1}(1/2)$?

Yes, absolutely! Remember the fundamental identity $\tan^{-1}(x) + \cot^{-1}(x) = \pi/2$. Therefore, $\pi/2 - \tan^{-1}(1/2)$ is exactly $\cot^{-1}(1/2)$, which is also equal to $\tan^{-1}(2)$.

Question

The sum of the infinite series $\cot^{-1} \left( \frac{7}{4} ight) + \cot^{-1} \left( \frac{19}{4} ight) + \cot^{-1} \left( \frac{39}{4} ight) + \cot^{-1} \left( \frac{67}{4} ight) + \dots$ is :-

Options:

(A) $\frac{\pi}{2} + 	an^{-1} \left( \frac{1}{2} ight)$

(B) $\frac{\pi}{2} - \cot^{-1} \left( \frac{1}{2} ight)$

(C) $\frac{\pi}{2} + \cot^{-1} \left( \frac{1}{2} ight)$

(D) $\frac{\pi}{2} - 	an^{-1} \left( \frac{1}{2} ight)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic jumble of inverse trigonometric functions.

You see a series: $\cot^{-1} \left( \frac{7}{4} ight) + \cot^{-1} \left( \frac{19}{4} ight) + \cot^{-1} \left( \frac{39}{4} ight) + \dots$. Beneath this complexity lies a hidden, rhythmic order waiting for us to uncover it.

### Decoding the Pattern

Every great journey begins with observation. Let us look at the numerators: $7, 19, 39, 67, \dots$.

If we calculate the first differences, we get $12, 20, 28, \dots$. The second differences are $8, 8, \dots$, which confirms that the sequence is quadratic.

By fitting this to the form $an^2 + bn + c$, we discover that the $n$-th numerator is $N_n = 4n^2 + 3$. Thus, our general term is:

$$T_n = \cot^{-1} \left( \frac{4n^2 + 3}{4} ight)$$

### The Art of Transformation

We prefer working with $	an^{-1}$ because it facilitates the use of the difference identity. Using the property $\cot^{-1}(x) = 	an^{-1}\left(\frac{1}{x}ight)$, we rewrite our term as:

$$T_n = 	an^{-1} \left( \frac{4}{4n^2 + 3} ight)$$

Dividing the numerator and denominator by $4$, we obtain:

$$T_n = 	an^{-1} \left( \frac{1}{n^2 + 3/4} ight)$$

To utilize the identity $	an^{-1}(A) - 	an^{-1}(B)$, we force the denominator into the form $1 + AB$. We add and subtract $1$ in the denominator:

$$T_n = 	an^{-1} \left( \frac{1}{1 + (n^2 - 1/4)} ight)$$

Recognizing $n^2 - 1/4$ as a difference of squares, we factor it into $(n - 1/2)(n + 1/2)$. Our term now stands as:

$$T_n = 	an^{-1} \left( \frac{1}{1 + (n + 1/2)(n - 1/2)} ight)$$

### The Telescoping Symphony

Observe that the numerator can be expressed as the difference of our factors: $(n + 1/2) - (n - 1/2) = 1$. We can now express our term as:

$$T_n = 	an^{-1} \left( \frac{(n + 1/2) - (n - 1/2)}{1 + (n + 1/2)(n - 1/2)} ight)$$

By the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1} \left( \frac{A - B}{1 + AB} ight)$, this simplifies to:

$$T_n = 	an^{-1} \left( n + 1/2 ight) - 	an^{-1} \left( n - 1/2 ight)$$

When we sum these terms from $n=1$ to $N$, the terms cancel like falling dominoes. This leaves us with only the first negative term and the last positive term:

$$S_N = 	an^{-1} \left( N + 1/2 ight) - 	an^{-1} \left( 1/2 ight)$$

### Final Calculation

Finally, we take the limit as $N 	o \infty$. As $N$ grows, $	an^{-1} \left( N + 1/2 ight)$ approaches the horizontal asymptote of $\pi/2$.

Our infinite sum becomes:

$$S_\infty = \frac{\pi}{2} - 	an^{-1} \left( \frac{1}{2} ight)$$

Using the identity $\frac{\pi}{2} - 	an^{-1}(x) = \cot^{-1}(x)$, we arrive at the final result:

**$$S_\infty = \cot^{-1} \left( \frac{1}{2} ight)$$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: The sum of the infinite series $cot^{- 1} (\frac{7}{4}) + cot^{- 1} (\frac{19}{4}) + cot^{- 1} (\frac{39}{4}) + cot^{- 1} (\frac{67}{4}) + \dots$ is :-

Select Answer:

Visualized Solution

Analyzing the Series

Finding the Pattern

General Term $T_{n}$

Converting to $tan^{- 1}$

Simplifying the Argument

The Target Formula

Creating the $1 + A B$ Form

Factoring the Denominator

Matching the Numerator

Applying the Difference Formula

Writing the Partial Sum $S_{n}$

Telescoping Cancellation

Finding the Infinite Sum $S_{\infty}$

Final Answer

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

Decoding the Pattern

The Art of Transformation

The Telescoping Symphony

Final Calculation

Similar Questions

Let $S_{k} = \sum_{r = 1}^{k} tan^{- 1} (\frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}})$ . Then $lim_{k \to \infty} S_{k}$ equal to :

The value of $cot (\sum_{n = 1}^{19} cot^{- 1} (1 + \sum_{p = 1}^{n} 2 p))$ is :

For $n \in N$ , if $cot^{- 1} 3 + cot^{- 1} 4 + cot^{- 1} 5 + cot^{- 1} n = \frac{π}{4}$ , then $n$ is equal to

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - \frac{( a ^{2} + b ^{2} )}{2} + \frac{( a ^{3} + b ^{3} )}{3} - \frac{( a ^{4} + b ^{4} )}{4} + \dots$ is :

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

If $sin^{- 1} \frac{α}{17} + cos^{- 1} \frac{4}{5} - tan^{- 1} \frac{77}{36} = 0, 0 < α < 13$ , then $sin^{- 1} (sin α) + cos^{- 1} (cos α)$ is equal to

Let $S = {x : cos^{- 1} x = π + sin^{- 1} x + sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} (2 x - 1)^{2}$ is equal to

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the infinite series cot−1(47​)+cot−1(419​)+cot−1(439​)+cot−1(467​)+… is :-

Select Answer:

Visualized Solution

Analyzing the Series

Finding the Pattern

General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Converting to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1

Simplifying the Argument

The Target Formula

Creating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+AB Form