Question

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

Answer 1

0

Answer 2

Let the given limit be $L$. The expression is $L = \lim_{n \to \infty} n! \left( \prod_{K=1}^{n} \tan^{-1} \frac{1}{1+K+K^2} \right)$

Let the term inside the product be $a_K = \tan^{-1} \frac{1}{1+K+K^2}$. 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 $xy = K(K+1)$. So, $\tan^{-1}(K+1) - \tan^{-1} K = \tan^{-1} \frac{(K+1)-K}{1+(K+1)K} = \tan^{-1} \frac{1}{1+K+K^2}$. Thus, $a_K = \tan^{-1}(K+1) - \tan^{-1} K$.

The product is $P_n = \prod_{K=1}^{n} a_K = \prod_{K=1}^{n} (\tan^{-1}(K+1) - \tan^{-1} K)$. We need to evaluate the limit of $n! P_n$ as $n \to \infty$. Consider the logarithm of the expression $n! P_n$: $\ln(n! P_n) = \ln(n!) + \ln P_n = \ln(n!) + \sum_{K=1}^{n} \ln a_K$.

For large $K$, $a_K = \tan^{-1}(K+1) - \tan^{-1} K$ is small. We can use the Taylor expansion of $\tan^{-1} x$ for large $x$: $\tan^{-1} x = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \dots$. $a_K = \left(\frac{\pi}{2} - \frac{1}{K+1} + \frac{1}{3(K+1)^3} - \dots \right) - \left(\frac{\pi}{2} - \frac{1}{K} + \frac{1}{3K^3} - \dots \right)$ $a_K = \left(\frac{1}{K} - \frac{1}{K+1}\right) - \frac{1}{3}\left(\frac{1}{K^3} - \frac{1}{(K+1)^3}\right) + \dots$ $a_K = \frac{1}{K(K+1)} - \frac{1}{3} \frac{3K^2+3K+1}{K^3(K+1)^3} + \dots$ $a_K = \frac{1}{K^2+K} - \frac{3K^2+3K+1}{3(K^6+3K^5+3K^4+K^3)} + \dots$ For large $K$, $a_K = \frac{1}{K^2(1+1/K)} - \frac{3K^2(1+1/K+1/(3K^2))}{3K^6(1+3/K+3/K^2+1/K^3)} + \dots$ $a_K = \frac{1}{K^2}(1 - 1/K + O(1/K^2)) - \frac{1}{K^4}(1+O(1/K)) + \dots$ $a_K = \frac{1}{K^2} - \frac{1}{K^3} + O(\frac{1}{K^4})$.

Now consider $\ln a_K$ for large $K$: $\ln a_K = \ln \left( \frac{1}{K^2} - \frac{1}{K^3} + O(\frac{1}{K^4}) \right) = \ln \left( \frac{1}{K^2} (1 - 1/K + O(1/K^2)) \right)$ $\ln a_K = \ln(1/K^2) + \ln(1 - 1/K + O(1/K^2))$ $\ln a_K = -2 \ln K + (-1/K + O(1/K^2)) - \frac{1}{2}(-1/K + O(1/K^2))^2 + O((-1/K)^3)$ $\ln a_K = -2 \ln K - 1/K + O(1/K^2)$.

Now sum $\ln a_K$ from $K=1$ to $n$: $\sum_{K=1}^{n} \ln a_K = \sum_{K=1}^{n} (-2 \ln K - 1/K + O(1/K^2))$ $\sum_{K=1}^{n} \ln a_K = -2 \sum_{K=1}^{n} \ln K - \sum_{K=1}^{n} 1/K + \sum_{K=1}^{n} O(1/K^2)$ $\sum_{K=1}^{n} \ln a_K = -2 \ln(n!) - H_n + C_n$, where $H_n = \sum_{K=1}^{n} 1/K$ is the $n$-th harmonic number and $C_n = \sum_{K=1}^{n} O(1/K^2)$. As $n \to \infty$, $C_n$ converges to a constant $C = \sum_{K=1}^{\infty} O(1/K^2)$.

Now substitute this back into the expression for $\ln(n! P_n)$: $\ln(n! P_n) = \ln(n!) + (-2 \ln(n!) - H_n + C_n) = -\ln(n!) - H_n + C_n$.

As $n \to \infty$, we use the asymptotic approximations: $\ln(n!) \approx n \ln n - n$ (by Stirling's formula) $H_n \approx \ln n + \gamma$, where $\gamma$ is the Euler-Mascheroni constant. $C_n \to C$.

So, $\lim_{n \to \infty} \ln(n! P_n) = \lim_{n \to \infty} (-\ln(n!) - H_n + C_n)$ $= \lim_{n \to \infty} (-(n \ln n - n + O(\ln n)) - (\ln n + \gamma + o(1)) + C + o(1))$ $= \lim_{n \to \infty} (-n \ln n + n - \ln n - \gamma + C + O(\ln n))$ The dominant term is $-n \ln n$, which tends to $-\infty$ as $n \to \infty$. $\lim_{n \to \infty} \ln(n! P_n) = -\infty$.

Therefore, the limit of the original expression is: $L = \lim_{n \to \infty} n! P_n = e^{\lim_{n \to \infty} \ln(n! P_n)} = e^{-\infty} = 0$.

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