Let Sk = Σ (r=1 to k) tan⁻¹(6ʳ / (2²ʳ⁺¹ + 3²ʳ⁺¹)). Then lim (k→∞) Sk equal to : (A) tan⁻¹(3/2) (B) π

Q: Why do we divide by $3^{2r+1}$? Is there a rule for this?

That is a fantastic question! In problems involving $\tan^{-1}$ summations, we almost always aim to use the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$. To use this, we need a '1' in the denominator. Dividing by the largest term in the denominator is a standard, powerful technique to force that '1' to appear. It is not just a rule; it is a strategic move to transform the expression into a form we can actually solve.

Q: How do I know what $x$ and $y$ should be?

Once you have the expression in the form $\frac{x-y}{1+xy}$, you can identify $x$ and $y$ by looking at the components. In this problem, after dividing by $3^{2r+1}$, the denominator becomes $1 + (2/3)^{2r+1}$. This looks like $1 + (2/3)^r \cdot (2/3)^{r+1}$. If you test $x = (2/3)^r$ and $y = (2/3)^{r+1}$, you will find that their difference $x-y$ perfectly matches your numerator. It is a bit of trial and error, but it becomes intuitive with practice!

Q: Why does the limit go to zero for the second term?

Remember that we are dealing with the term $(2/3)^{k+1}$. Since $2/3$ is a fraction less than 1, as $k$ approaches infinity, the exponent $k+1$ also approaches infinity. Any fraction between 0 and 1 raised to an infinitely large power will shrink to 0. Therefore, $\tan^{-1}(0)$ becomes 0, which is why that term vanishes.

Q: Why is the answer $\cot^{-1}(3/2)$ and not $\tan^{-1}(2/3)$?

You are absolutely right that the sum evaluates to $\tan^{-1}(2/3)$. However, in competitive exams, the options are often presented in different forms. We use the identity $\tan^{-1}(x) = \cot^{-1}(1/x)$ for $x > 0$. By applying this, $\tan^{-1}(2/3)$ becomes $\cot^{-1}(3/2)$. Always check if your answer can be rewritten to match the given options!

Question

Let $S_k = \sum_{r=1}^k 	an^{-1} \left( \frac{6^r}{2^{2r+1} + 3^{2r+1}} ight)$. Then $\lim_{k 	o \infty} S_k$ equal to :

Options:

(A) $	an^{-1} \left( \frac{3}{2} ight)$

(B) $\frac{\pi}{2}$

(C) $\cot^{-1} \left( \frac{3}{2} ight)$

(D) $	an^{-1}(3)$

Sigma Percentile · Accepted Answer

### The Symphony of Telescoping Series

Welcome, fellow traveler of the mathematical landscape. Today, we are going to embark on a journey through a problem that might look intimidating at first glance, but hides a beautiful, elegant structure.

We are tasked with finding the sum of an infinite series involving the inverse tangent function: 
$$S_{\infty} = \lim_{k 	o \infty} \sum_{r=1}^k 	an^{-1} \left( \frac{6^r}{2^{2r+1} + 3^{2r+1}} ight)$$

### Phase 1: The Anatomy of the Term

When you see a summation of $	an^{-1}$ terms, your intuition should immediately scream **'telescoping series!'** The goal in such problems is to break down the general term $T_r$ into a difference of two terms, say $	an^{-1}(x) - 	an^{-1}(y)$.

This allows the series to collapse like a telescope, where intermediate terms cancel each other out. Our general term is:
$$T_r = 	an^{-1} \left( \frac{6^r}{2^{2r+1} + 3^{2r+1}} ight)$$

### Phase 2: The Art of Manipulation

To use the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$, we need a '1' in the denominator. Looking at our denominator $2^{2r+1} + 3^{2r+1}$, the most natural way to create a '1' is to divide both the numerator and the denominator by the larger term, $3^{2r+1}$.

Let us perform this operation:
$$T_r = 	an^{-1} \left( \frac{\frac{6^r}{3^{2r+1}}}{\frac{2^{2r+1} + 3^{2r+1}}{3^{2r+1}}} ight)$$

Now, let us simplify the numerator. We know $6^r = 2^r \cdot 3^r$. So:
$$\frac{6^r}{3^{2r+1}} = \frac{2^r \cdot 3^r}{3^{2r} \cdot 3^1} = \frac{2^r}{3^r} \cdot \frac{1}{3} = \left( \frac{2}{3} ight)^r \cdot \frac{1}{3}$$

For the denominator, we get:
$$\frac{2^{2r+1}}{3^{2r+1}} + \frac{3^{2r+1}}{3^{2r+1}} = \left( \frac{2}{3} ight)^{2r+1} + 1$$
This is exactly the structure we needed.

### Phase 3: The Telescoping Symphony

Now, we need to find $x$ and $y$ such that $x-y = \left( \frac{2}{3} ight)^r \cdot \frac{1}{3}$ and $xy = \left( \frac{2}{3} ight)^{2r+1}$. If we set $x = \left( \frac{2}{3} ight)^r$ and $y = \left( \frac{2}{3} ight)^{r+1}$, notice that:
$$xy = \left( \frac{2}{3} ight)^r \cdot \left( \frac{2}{3} ight)^{r+1} = \left( \frac{2}{3} ight)^{2r+1}$$

And for the difference:
$$x-y = \left( \frac{2}{3} ight)^r - \left( \frac{2}{3} ight)^{r+1} = \left( \frac{2}{3} ight)^r \left( 1 - \frac{2}{3} ight) = \left( \frac{2}{3} ight)^r \cdot \frac{1}{3}$$
It fits perfectly!

So, the term simplifies to:
$$T_r = 	an^{-1} \left( \frac{\left( \frac{2}{3} ight)^r - \left( \frac{2}{3} ight)^{r+1}}{1 + \left( \frac{2}{3} ight)^r \cdot \left( \frac{2}{3} ight)^{r+1}} ight) = 	an^{-1} \left( \frac{2}{3} ight)^r - 	an^{-1} \left( \frac{2}{3} ight)^{r+1}$$

### Phase 4: The Final Limit

When we sum these terms from $r=1$ to $k$, we get:
$$S_k = \sum_{r=1}^k \left[ 	an^{-1} \left( \frac{2}{3} ight)^r - 	an^{-1} \left( \frac{2}{3} ight)^{r+1} ight]$$

Expanding this, we see the telescoping effect:
$$S_k = \left( 	an^{-1} \left( \frac{2}{3} ight)^1 - 	an^{-1} \left( \frac{2}{3} ight)^2 ight) + \dots + \left( 	an^{-1} \left( \frac{2}{3} ight)^k - 	an^{-1} \left( \frac{2}{3} ight)^{k+1} ight)$$

All intermediate terms cancel out, leaving:
$$S_k = 	an^{-1} \left( \frac{2}{3} ight) - 	an^{-1} \left( \frac{2}{3} ight)^{k+1}$$

As $k 	o \infty$, $\left( \frac{2}{3} ight)^{k+1} 	o 0$, so:
$$S_{\infty} = 	an^{-1} \left( \frac{2}{3} ight) - 	an^{-1}(0) = 	an^{-1} \left( \frac{2}{3} ight)$$

Finally, we convert this to $\cot^{-1}$ form:
$$	an^{-1} \left( \frac{2}{3} ight) = \cot^{-1} \left( \frac{3}{2} ight)$$

And there we have it—the elegance of the solution revealed! The final answer is **$\cot^{-1} \left( \frac{3}{2} ight)$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: 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 :

Select Answer:

Visualized Solution

Analyze the Summation $S_{k}$

The Difference Formula for $tan^{- 1}$

Normalizing the Expression

Simplifying the Numerator

Simplifying the Denominator

Identifying $x$ and $y$

The Telescoping Form

Expanding the Sum $S_{k}$

Cancellation of Terms

Applying the Limit $k \to \infty$

Final Result in $cot^{- 1}$

The Sigma Insight: Solving Inverse Trigonometric Equations

The Symphony of Telescoping Series

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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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; } Let Sk​=∑r=1k​tan−1(22r+1+32r+16r​). Then limk→∞​Sk​ equal to :

Select Answer:

Visualized Solution

Analyze the Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sk​

The Difference Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1

Normalizing the Expression

Simplifying the Numerator

Simplifying the Denominator

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

The Telescoping Form

Expanding the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sk​

Cancellation of Terms

Applying the Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k→∞

Final Result in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cot−1