Question

$\lim_{n\to\infty} \ln\prod_{K=1}^{n} \tan^{-1}(\frac{1}{1+K+K^2})$

Answer 1

-\infty

Answer 2

Let the given limit be $L$. $L = \lim_{n\to\infty} \ln\prod_{K=1}^{n} \tan^{-1}\left(\frac{1}{1+K+K^2}\right)$ This can be rewritten as a sum: $L = \lim_{n\to\infty} \sum_{K=1}^{n} \ln\left(\tan^{-1}\left(\frac{1}{1+K+K^2}\right)\right)$ Let $a_K = \tan^{-1}\left(\frac{1}{1+K+K^2}\right)$. We use the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$. Let $x = K+1$ and $y = K$. Then $x-y = (K+1)-K = 1$ and $xy = K(K+1) = K^2+K$. So, $a_K = \tan^{-1}(K+1) - \tan^{-1} K$.

Now, we analyze the behavior of $a_K$ for large $K$. Using the Taylor expansion of $\tan^{-1} x$ for large $x$, we have $\tan^{-1} x = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - O(\frac{1}{x^5})$. So, for large $K$: $a_K = \left(\frac{\pi}{2} - \frac{1}{K+1} + \frac{1}{3(K+1)^3} - \dots\right) - \left(\frac{\pi}{2} - \frac{1}{K} + \frac{1}{3K^3} - \dots\right)$ $a_K = \left(\frac{1}{K} - \frac{1}{K+1}\right) - \frac{1}{3}\left(\frac{1}{K^3} - \frac{1}{(K+1)^3}\right) + \dots$ $a_K = \frac{1}{K(K+1)} - \frac{1}{3} \frac{(K+1)^3 - K^3}{K^3(K+1)^3} + \dots$ $a_K = \frac{1}{K^2+K} - \frac{1}{3} \frac{3K^2+3K+1}{K^3(K+1)^3} + \dots$ For large $K$, the dominant term is $\frac{1}{K^2+K} = \frac{1}{K^2(1+1/K)} = \frac{1}{K^2}(1 - \frac{1}{K} + O(\frac{1}{K^2})) = \frac{1}{K^2} - \frac{1}{K^3} + O(\frac{1}{K^4})$. So, $a_K = \frac{1}{K^2} - \frac{1}{K^3} + O(\frac{1}{K^4})$.

Now consider the logarithm of $a_K$: $\ln(a_K) = \ln\left(\frac{1}{K^2} - \frac{1}{K^3} + O(\frac{1}{K^4})\right)$ $\ln(a_K) = \ln\left(\frac{1}{K^2}\left(1 - \frac{1}{K} + O(\frac{1}{K^2})\right)\right)$ $\ln(a_K) = \ln(K^{-2}) + \ln\left(1 - \frac{1}{K} + O(\frac{1}{K^2})\right)$ Using the Taylor expansion $\ln(1+u) = u - \frac{u^2}{2} + O(u^3)$ for small $u$, with $u = -\frac{1}{K} + O(\frac{1}{K^2})$: $\ln(a_K) = -2 \ln K + \left(-\frac{1}{K} + O(\frac{1}{K^2})\right) - \frac{1}{2}\left(-\frac{1}{K} + O(\frac{1}{K^2})\right)^2 + O\left(\left(-\frac{1}{K}\right)^3\right)$ $\ln(a_K) = -2 \ln K - \frac{1}{K} + O(\frac{1}{K^2})$.

Now we sum $\ln(a_K)$ from $K=1$ to $n$: $\sum_{K=1}^{n} \ln(a_K) = \sum_{K=1}^{n} \left(-2 \ln K - \frac{1}{K} + O(\frac{1}{K^2})\right)$ $= -2 \sum_{K=1}^{n} \ln K - \sum_{K=1}^{n} \frac{1}{K} + \sum_{K=1}^{n} O(\frac{1}{K^2})$

We know that $\sum_{K=1}^{n} \ln K = \ln(n!)$ and $\sum_{K=1}^{n} \frac{1}{K} = H_n$ (the $n$-th harmonic number). The sum $\sum_{K=1}^{n} O(\frac{1}{K^2})$ converges to a finite constant as $n \to \infty$ because $\sum \frac{1}{K^2}$ converges. Let this sum be $C_n$, which tends to a constant $C$ as $n \to \infty$. So, $\sum_{K=1}^{n} \ln(a_K) = -2 \ln(n!) - H_n + C_n$.

We need to find the limit of this expression as $n \to \infty$. Using Stirling's approximation for $\ln(n!)$: $\ln(n!) \approx n \ln n - n$ for large $n$. Using the approximation for $H_n$: $H_n \approx \ln n + \gamma$ for large $n$, where $\gamma$ is the Euler-Mascheroni constant.

Substituting these into the sum: $\lim_{n\to\infty} \sum_{K=1}^{n} \ln(a_K) = \lim_{n\to\infty} (-2 \ln(n!) - H_n + C_n)$ $\approx \lim_{n\to\infty} (-2(n \ln n - n) - (\ln n + \gamma) + C)$ $= \lim_{n\to\infty} (-2n \ln n + 2n - \ln n - \gamma + C)$

The dominant term in this expression is $-2n \ln n$. As $n \to \infty$, $-2n \ln n \to -\infty$. Therefore, the limit of the sum is $-\infty$. $L = \lim_{n\to\infty} \sum_{K=1}^{n} \ln(a_K) = -\infty$

The question asks for the limit of the logarithm of the product, which is exactly this sum.