Question

If $\tan^{-1}(\frac{1}{2.1^2})+\tan^{-1}(\frac{1}{2.2^2})+\tan^{-1}(\frac{1}{2.3^2})+..$ up to infinite terms = $\lambda$.

Then the value of $[\tan \frac{\lambda}{2}]$, where [.] represents the greatest integer function, is

Answer 1

0

Answer 2

The given infinite series is $S = \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2n^2}\right)$. Let the general term be $T_n = \tan^{-1}\left(\frac{1}{2n^2}\right)$.

We want to express $T_n$ in the form $\tan^{-1}(A) - \tan^{-1}(B)$, using the identity: $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$.

We need to find $x$ and $y$ such that $\frac{x-y}{1+xy} = \frac{1}{2n^2}$. Let's try to set $x-y = 1$ and $1+xy = 2n^2$. This approach did not lead to simple terms for $x$ and $y$.

Let's try to manipulate the argument $\frac{1}{2n^2}$ by multiplying the numerator and denominator by a constant, say 2: $\frac{1}{2n^2} = \frac{2}{4n^2}$. Now we want to express $\frac{2}{4n^2}$ in the form $\frac{x-y}{1+xy}$. Let $x-y=2$ and $1+xy=4n^2$. This means $xy = 4n^2-1$. We are looking for two terms $x$ and $y$ such that their difference is 2 and their product is $4n^2-1$. Notice that $4n^2-1 = (2n)^2 - 1^2 = (2n-1)(2n+1)$. So, if we choose $x=2n+1$ and $y=2n-1$, then: $x-y = (2n+1) - (2n-1) = 2$. $xy = (2n+1)(2n-1) = 4n^2-1$. This works perfectly for the argument $\frac{2}{4n^2}$.

However, our original term is $\tan^{-1}\left(\frac{1}{2n^2}\right)$, not $\tan^{-1}\left(\frac{2}{4n^2}\right)$. Let's try to choose $x$ and $y$ such that $x-y=1$ and $1+xy=2n^2$. This was the first attempt. Let $x = \frac{1}{2n-1}$ and $y = \frac{1}{2n+1}$. Then, $x-y = \frac{1}{2n-1} - \frac{1}{2n+1} = \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)} = \frac{2}{4n^2-1}$. $1+xy = 1 + \frac{1}{(2n-1)(2n+1)} = 1 + \frac{1}{4n^2-1} = \frac{4n^2-1+1}{4n^2-1} = \frac{4n^2}{4n^2-1}$. Now, substituting these into the $\tan^{-1}$ identity: $\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}\left(\frac{\frac{2}{4n^2-1}}{\frac{4n^2}{4n^2-1}}\right) = \tan^{-1}\left(\frac{2}{4n^2}\right) = \tan^{-1}\left(\frac{1}{2n^2}\right)$. This is exactly the general term $T_n$.

So, we have found that: $T_n = \tan^{-1}\left(\frac{1}{2n^2}\right) = \tan^{-1}\left(\frac{1}{2n-1}\right) - \tan^{-1}\left(\frac{1}{2n+1}\right)$.

Now, let's write out the sum for the first $N$ terms, $S_N$: For $n=1$: $T_1 = \tan^{-1}\left(\frac{1}{1}\right) - \tan^{-1}\left(\frac{1}{3}\right)$ For $n=2$: $T_2 = \tan^{-1}\left(\frac{1}{3}\right) - \tan^{-1}\left(\frac{1}{5}\right)$ For $n=3$: $T_3 = \tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{7}\right)$ ... For $n=N$: $T_N = \tan^{-1}\left(\frac{1}{2N-1}\right) - \tan^{-1}\left(\frac{1}{2N+1}\right)$

Summing these terms, we see that it's a telescoping series: $S_N = \sum_{n=1}^{N} \left[ \tan^{-1}\left(\frac{1}{2n-1}\right) - \tan^{-1}\left(\frac{1}{2n+1}\right) \right] = \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{2N+1}\right)$.

The problem asks for the sum up to infinite terms, which is $\lambda$. So we take the limit as $N \to \infty$: $\lambda = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left[ \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{2N+1}\right) \right]$. As $N \to \infty$, the term $\frac{1}{2N+1} \to 0$. So, $\tan^{-1}\left(\frac{1}{2N+1}\right) \to \tan^{-1}(0) = 0$. Therefore, $\lambda = \tan^{-1}(1) - 0 = \frac{\pi}{4}$.

Finally, we need to find the value of $[\tan \frac{\lambda}{2}]$. Substitute $\lambda = \frac{\pi}{4}$: $\frac{\lambda}{2} = \frac{\pi/4}{2} = \frac{\pi}{8}$. Now, we need to calculate $\tan\left(\frac{\pi}{8}\right)$. We know the half-angle formula for tangent: $\tan\left(\frac{\theta}{2}\right) = \frac{1-\cos\theta}{\sin\theta}$. Let $\theta = \frac{\pi}{4}$. $\tan\left(\frac{\pi}{8}\right) = \frac{1-\cos(\pi/4)}{\sin(\pi/4)}$. We know $\cos(\pi/4) = \frac{1}{\sqrt{2}}$ and $\sin(\pi/4) = \frac{1}{\sqrt{2}}$. $\tan\left(\frac{\pi}{8}\right) = \frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2}-1$.

Now, we need to find the greatest integer function of $\tan\left(\frac{\pi}{8}\right)$: $[\tan \frac{\lambda}{2}] = [\sqrt{2}-1]$. We know that $\sqrt{2} \approx 1.414$. So, $\sqrt{2}-1 \approx 1.414 - 1 = 0.414$. The greatest integer less than or equal to $0.414$ is $0$.

The final answer is $\boxed{0}$.

Explanation of the solution:

1. Identify the general term: The given series is $\sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2n^2}\right)$. Let $T_n = \tan^{-1}\left(\frac{1}{2n^2}\right)$.
2. Apply telescoping sum technique: The goal is to express $T_n$ in the form $\tan^{-1}(f(n)) - \tan^{-1}(f(n+1))$ or similar. Use the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. By choosing $x = \frac{1}{2n-1}$ and $y = \frac{1}{2n+1}$, we find that: $\frac{x-y}{1+xy} = \frac{\frac{1}{2n-1} - \frac{1}{2n+1}}{1 + \frac{1}{(2n-1)(2n+1)}} = \frac{\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}}{\frac{(2n-1)(2n+1)+1}{(2n-1)(2n+1)}} = \frac{2}{4n^2-1+1} = \frac{2}{4n^2} = \frac{1}{2n^2}$. Thus, $T_n = \tan^{-1}\left(\frac{1}{2n-1}\right) - \tan^{-1}\left(\frac{1}{2n+1}\right)$.
3. Sum the series: Write out the first few terms of the series to observe the telescoping pattern: $T_1 = \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{3}\right)$ $T_2 = \tan^{-1}\left(\frac{1}{3}\right) - \tan^{-1}\left(\frac{1}{5}\right)$ $T_3 = \tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{7}\right)$ ... The sum of the first $N$ terms is $S_N = \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{2N+1}\right)$.
4. Find the infinite sum ($\lambda$): Take the limit as $N \to \infty$: $\lambda = \lim_{N \to \infty} \left( \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{2N+1}\right) \right) = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.
5. Calculate the final expression: The question asks for $[\tan \frac{\lambda}{2}]$. $\frac{\lambda}{2} = \frac{\pi/4}{2} = \frac{\pi}{8}$. Calculate $\tan\left(\frac{\pi}{8}\right)$ using the half-angle formula $\tan\left(\frac{\theta}{2}\right) = \frac{1-\cos\theta}{\sin\theta}$ with $\theta = \frac{\pi}{4}$: $\tan\left(\frac{\pi}{8}\right) = \frac{1-\cos(\pi/4)}{\sin(\pi/4)} = \frac{1-1/\sqrt{2}}{1/\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}} \cdot \sqrt{2} = \sqrt{2}-1$.
6. Apply the greatest integer function: $[\tan \frac{\lambda}{2}] = [\sqrt{2}-1]$. Since $\sqrt{2} \approx 1.414$, $\sqrt{2}-1 \approx 0.414$. The greatest integer less than or equal to $0.414$ is $0$.