Question

Let $a > 0$ and $f: (-\infty, -(a+1)) \cup (-a, +\infty) \rightarrow \mathbb{R},$ $f(x) = \frac{1}{x^2 + (2a+1)x + a^2 + a}$ Determine the value of $\lim_{n \to \infty} \sqrt[n^2]{\left| \lim_{p \to \infty} \sum_{k=1}^{p} f^n(k) \right|}$ where $f^n(x)$ denotes the $n^{th}$ derivative of $f$.

Answer 1

1

Answer 2

The function $f(x)$ can be simplified by factoring the denominator: $x^2 + (2a+1)x + a^2 + a = (x+a)(x+a+1)$ So, $f(x) = \frac{1}{(x+a)(x+a+1)}$ Using partial fraction decomposition, we get: $f(x) = \frac{1}{x+a} - \frac{1}{x+a+1}$ The $n^{th}$ derivative of $f(x)$ is: $f^{(n)}(x) = \frac{d^n}{dx^n}\left(\frac{1}{x+a}\right) - \frac{d^n}{dx^n}\left(\frac{1}{x+a+1}\right)$ The $n^{th}$ derivative of $(x+c)^{-1}$ is $\frac{(-1)^n n!}{(x+c)^{n+1}}$. Thus, $f^{(n)}(x) = (-1)^n n! \left( \frac{1}{(x+a)^{n+1}} - \frac{1}{(x+a+1)^{n+1}} \right)$ Now, consider the inner limit: $\lim_{p \to \infty} \sum_{k=1}^{p} f^n(k) = \lim_{p \to \infty} \sum_{k=1}^{p} (-1)^n n! \left( \frac{1}{(k+a)^{n+1}} - \frac{1}{(k+a+1)^{n+1}} \right)$ This is a telescoping sum: $= (-1)^n n! \lim_{p \to \infty} \left[ \left(\frac{1}{(1+a)^{n+1}} - \frac{1}{(2+a)^{n+1}}\right) + \left(\frac{1}{(2+a)^{n+1}} - \frac{1}{(3+a)^{n+1}}\right) + \dots + \left(\frac{1}{(p+a)^{n+1}} - \frac{1}{(p+1+a)^{n+1}}\right) \right]$ $= (-1)^n n! \lim_{p \to \infty} \left( \frac{1}{(1+a)^{n+1}} - \frac{1}{(p+1+a)^{n+1}} \right)$ As $p \to \infty$, $\frac{1}{(p+1+a)^{n+1}} \to 0$. So, the limit of the sum is: $(-1)^n n! \frac{1}{(1+a)^{n+1}}$ Now, we take the absolute value: $\left| (-1)^n n! \frac{1}{(1+a)^{n+1}} \right| = \frac{n!}{(1+a)^{n+1}}$ Finally, we evaluate the outer limit: $L = \lim_{n \to \infty} \sqrt[n^2]{\frac{n!}{(1+a)^{n+1}}} = \lim_{n \to \infty} \left( \frac{n!}{(1+a)^{n+1}} \right)^{\frac{1}{n^2}}$ To evaluate this, we consider the logarithm: $\ln L = \lim_{n \to \infty} \frac{1}{n^2} \ln \left( \frac{n!}{(1+a)^{n+1}} \right) = \lim_{n \to \infty} \frac{\ln(n!) - (n+1)\ln(1+a)}{n^2}$ We use the fact that $\ln(n!) \approx n \ln n - n$ (Stirling's approximation). $\ln L = \lim_{n \to \infty} \frac{n \ln n - n - (n+1)\ln(1+a)}{n^2}$ $\ln L = \lim_{n \to \infty} \left( \frac{n \ln n}{n^2} - \frac{n}{n^2} - \frac{(n+1)\ln(1+a)}{n^2} \right)$ $\ln L = \lim_{n \to \infty} \left( \frac{\ln n}{n} - \frac{1}{n} - \frac{(n+1)\ln(1+a)}{n^2} \right)$ As $n \to \infty$: $\lim_{n \to \infty} \frac{\ln n}{n} = 0$ (by L'Hopital's rule). $\lim_{n \to \infty} \frac{1}{n} = 0$. $\lim_{n \to \infty} \frac{(n+1)\ln(1+a)}{n^2} = \ln(1+a) \lim_{n \to \infty} \frac{n+1}{n^2} = \ln(1+a) \cdot 0 = 0$. Therefore, $\ln L = 0 - 0 - 0 = 0$. Since $\ln L = 0$, we have $L = e^0 = 1$.