Binomial Logarithms

We prove by induction that $J_n(a) \; = \; \int_0^1 \int_0^1 \cdots \int_0^1 \frac{1}{x_1+x_2+\cdots + x_n + a}\,dx_1\,dx_2\,\cdots\,dx_n \; = \; \frac{1}{(n-1)!}\sum_{j=0}^n {n \choose j}(-1)^{n-j}(a+j)^{n-1} \ln(a+j)$ for all $n \in \mathbb{N}$ and $a > 0$ . To start with, $J_1(a) \; = \; \int_0^1 \frac{dx}{x+a} \; = \; \ln(a+1) - \ln a$ If we assume that the formula is correct for $J_n(a)$ , then $\begin{aligned} J_{n+1}(a) & = \int_0^1 J_n(a+x)\,dx \; = \; \int_0^1 \left( \frac{1}{(n-1)!}\sum_{j=0}^n {n \choose j}(-1)^{n-j}(a+x+j)^{n-1}\ln(a+x+j)\right)\,dx \\ & = \frac{1}{(n-1)!}\sum_{j=0}^n {n \choose j}(-1)^{n-j} \left[\tfrac{1}{n}(a+x+j)^n\ln(a+x+j) - \tfrac{1}{n^2}(a+x+j)^n\right]_{x=0}^1 \\ & = \frac{1}{n!} \sum_{j=0}^{n+1}{n+1 \choose j}(-1)^{n+1-j}(a+j)^n \ln(a+j) - \frac{1}{n\times n!}\sum_{j=0}^{n+1}{n+1 \choose j}(-1)^{n+1-j}(a+j)^n \end{aligned}$ Now $\sum_{j=0}^N {N \choose j}(-1)^j (N-j)^M \; = \; \sum_{j=0}^N {N \choose j}(-1)^{N-j} j^M$ is equal to the number of surjective maps from a set with $M$ elements to a set with $N$ elements, and hence is equal to $0$ for non-negative integers $M < N$ . Thus it follows that $\sum_{j=0}^{n+1}{n+1 \choose j}(-1)^{n+1-j}(a+j)^n \; = \; 0$ for all $a > 0$ , and hence $J_{n+1}(a) \; = \; \frac{1}{n!} \sum_{j=0}^{n+1}{n+1 \choose j}(-1)^{n+1-j}(a+j)^n \ln(a+j)$ as required. Putting $a=0$ we conclude that $J_n(0) \; = \; \frac{1}{(n-1)!}\sum_{j=1}^n {n \choose j}(-1)^{n-j} j^{n-1} \ln j \; = \; n \sum_{j=1}^n \frac{(-1)^{n-j} j^{n-1} \ln j}{j! (n-j)!} \hspace{1cm} n \ge 2$

As observed elsewhere , we note that $n\,J_n(0) \; =\; \int_0^\infty e^{-\frac12t} \left(\frac{2n\sinh \frac1{2n}t}{t}\right)^n\,dt$ for all $n \in \mathbf{N}$ . Moreover, $\lim_{n \to \infty}n\,J_n(0) \; = \; \int_0^\infty e^{-\frac12t}\,dt \; = \; 2 \hspace{4cm} (\star)$ Putting these results together, we deduce that $\lim_{n \to \infty} n^2\sum_{j=0}^n \frac{(-1)^{n-j} j^{n-1} \ln j}{j! (n-j)!} \; =\; \lim_{n \to \infty} n\,J_n(0) \; = \; 2$ making the answer $\boxed{2}$ .

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Proving the result in $(\star)$ that $n\,J_n(0)$ converges to $2$ as $n \to \infty$ needs some careful analysis. If we define $g(x) \; = \; x\mathrm{coth}\,x - 1 - \ln\left(\frac{\sinh x}{x}\right) \hspace{2cm} x > 0$ then $g(x) \; =\; \cosh x \times \frac{x}{\sinh x} - 1 - \ln\left(\frac{\sinh x}{x}\right) \; \to \; 0 \hspace{2cm} x \to 0+$ while $g'(x) \; =\; \frac{1}{x} - x\mathrm{cosech}^2x \; = \; \frac{1}{x}\left[1 - \left(\frac{x}{\sinh x}\right)^2\right] \; > \; 0 \hspace{2cm} x > 0$ so we deduce that $g(x) > 0$ for all $x > 0$ . If we now define $f(x) \; = \; \left(\frac{\sinh x}{x}\right)^{\frac{1}{x}} \hspace{2cm} x > 0$ then we deduce that $\frac{f'(x)}{f(x)} \; = \; \frac{1}{x}\Big[\mathrm{coth}\,x - \frac{1}{x}\Big] - \frac{1}{x^2}\ln\left(\frac{\sinh x}{x}\right) \; = \; \frac{g(x)}{x^2} \; > \; 0$ for $x > 0$ . Thus $f(x)$ is an increasing function of $x$ . Thus, for $n \ge 2$ and $t > 0$ , $\begin{aligned} f\big(\tfrac{t}{2n}\big) & \le f\big(\tfrac{t}{4}\big) \\ \left(\frac{2n \sinh \frac{1}{2n}t}{t}\right)^{\frac{2n}{t}} & \le \left(\frac{4\sinh \frac14t}{t}\right)^{\frac4t} \\ e^{-\frac12t}\left(\frac{2n \sinh \frac{1}{2n}t}{t}\right)^n & \le e^{-\frac12t}\left(\frac{4\sinh \frac14t}{t}\right)^2 \; = \; \frac{(1 - e^{-t})^2}{t^2} \end{aligned}$ Given this inequality, we can use the Dominated Convergence Theorem to show that $\lim_{n \to \infty}n\,J_n(0) = 2$ .