An irresistible limit

Calculus (Integrals, Series)
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Tolaso J Kos
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An irresistible limit

#1

Post by Tolaso J Kos »

Evaluate the limit:

$$ \lim_{n \to +\infty} \dfrac{(-1)^nn^2}{n!} \sum\limits_{k=2}^{n}\binom{n}{k}(-1)^kk^{n-1}\ln k $$
Answer
The final answer of the limit is $2$.
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r9m
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Re: An irresistible limit

#2

Post by r9m »

Hey T, thanks for sharing my problem! :) Here's the original motivation behind creation of the problem:

Consider, the partial fraction decomposition of, $\displaystyle f_n(x) = \frac{x^{m}}{\prod\limits_{k=1}^{n}(x+k)} = \sum\limits_{k=1}^{n} \frac{A_k}{x+k}$ for some $m < n$, then for each $1 \le k \le n$, we have \begin{align*} A_k = \lim\limits_{x \to -k} (x+k)f_n(x) &= \lim\limits_{x\to -k} \frac{x^m}{\prod\limits_{j=1}^{k-1} (x+j) \prod\limits_{j=k+1}^{n} (x+j)} \\&= \frac{(-1)^{m-k+1} k^m}{(k-1)! (n-k)!} \\&= \frac{(-1)^{m-k+1}}{n!} \binom{n}{k}k^{m+1}\end{align*}

In particular when $m = n-2$ we have, $\displaystyle \frac{x^{n-2}}{\prod\limits_{k=1}^{n}(x+k)} = \frac{(-1)^n}{n!} \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n}{k}\frac{k^{n-1}}{x+k}$. Thus, \begin{align*}\int_0^{\infty} \frac{x^{n-2}}{\prod\limits_{k=1}^{n}(x+k)}\,dx &= \frac{(-1)^n}{n!} \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n}{k}k^{n-1}\int_0^{\infty}\frac{1}{x+k}\,dx \\&= \frac{(-1)^{n}}{n!} \sum\limits_{k=1}^{n} (-1)^{k}\binom{n}{k}k^{n-1}\log k\end{align*}

Now, using the partial fraction decomposition of $\displaystyle \frac{1}{\prod\limits_{k=0}^{n}(x+k)}$ we have, \begin{align*} \int_0^{\infty} \frac{x^{n-2}}{\prod\limits_{k=1}^{n}(x+k)}\,dx &= \frac{1}{n!}\int_0^{\infty} x^{n-1}\sum\limits_{k=0}^{n} (-1)^{k}\binom{n}{k}\frac{1}{x+k}\,dx\\&= \frac{1}{n!}\int_0^{\infty} x^{n-1}\sum\limits_{k=0}^{n} (-1)^{k}\binom{n}{k}\int_0^{\infty} e^{-(x+k)s}\,ds\,dx\\&= \frac{1}{n!} \int_0^{\infty}\int_0^{\infty} x^{n-1} (1-e^{-s})^ne^{-xs}\,ds\,dx\\&= \frac{1}{n}\int_0^{\infty} \left(\frac{1-e^{-s}}{s}\right)^n\,ds\end{align*}

That is, \begin{align*}I_n = \frac{(-1)^{n}n^2}{n!} \sum\limits_{k=1}^{n} (-1)^{k}\binom{n}{k}k^{n-1}\log k &= n\int_0^{\infty} \left(\frac{1-e^{-s}}{s}\right)^n\,ds\end{align*}

Now, the function $\displaystyle f(s) = \frac{1-e^{-s}}{s}$ is monotone decreasing on $[0,\infty]$, with $\displaystyle f'(s) = \frac{(1+s)e^{-s} - 1}{s^2} < 0$ (since, $\displaystyle e^{-s} < \frac{1}{1+s}$). Hence, making the substitution $s \mapsto g(s)$, where, $g$ is the inverse of $f$, \begin{align*}I_n = -\int_0^1 ns^n\,d\left(g(s)\right) = -\frac{n}{n+1}\int_0^1 g'\left(s^{\frac{1}{n+1}}\right)\,ds\end{align*} thus, $\displaystyle \lim\limits_{n \to \infty} I_n = -g'(1) = -\frac{1}{f'(g(1))} = 2$.
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