Asymptotic formula/closed form of $\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}$

Question

I need an asymptotic formula/closed form for the sum $$\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}$$ where $n\in\mathbb{N}$

Denote $$S_n=\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}$$

$$S_n=\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+1) }{n+r+1}+ \sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(1+\frac{r}{n+1}) }{n+r+1} $$

By Wolfram alpha see here $$\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+1) }{n+r+1}= \frac{\sqrt{\pi}\ n!\ \log (n+1)}{2^{2n+1} \Gamma(n+\frac{3}{2})} $$ $$S_n= \frac{\sqrt{\pi}\ n!\ \log (n+1)}{2^{2n+1} \Gamma(n+\frac{3}{2})} + \sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(1+\frac{r}{n+1}) }{n+r+1} $$ Since $\frac{r}{n+1}<1$ so by expansion of logarithm, $$S_n= \frac{\sqrt{\pi}\ n!\ \log (n+1)}{2^{2n+1} \Gamma(n+\frac{3}{2})} + \sum_{r=0}^{n}\sum_{m=0}^\infty\frac{(-1)^r\binom{n}{r}(-1)^{m+1}(\frac{r}{n+1})^m }{m(n+r+1)} $$

I could not think of an idea to solve the problem. Any help will be highly appreciated.

Answer 1

$$S_n=\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}=-\frac{\partial }{\partial s}\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}}{(n+r+1)^s}\,\bigg|_{s=1}$$ Using $\,\displaystyle \frac{1}{(n+r+1)^s}=\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+r+1)}dt$ $$S_n=-\frac{\partial }{\partial s}\,\bigg|_{s=1}\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+1)}\left(\sum_{r=0}^n(-1)^r\binom{n}{r}e^{-rt}\right)dt$$ $$=-\frac{\partial }{\partial s}\,\bigg|_{s=1}\frac1{\Gamma(s)}\int_0^\infty t^{s-1}e^{-t(n+1)}(1-e^{-t})^ndt$$ $$\overset{x=e^{-t}}{=}-\gamma\int_0^1x^n(1-x)^ndx-\int_0^1\ln(-\ln x)x^n(1-x)^ndx$$ $$=-\gamma \,B(n+1;n+1)-\int_0^1\ln(-\ln x)x^n(1-x)^ndx$$ I'm not sure whether the last integral has a nice closed form, though its asymptotics at $n\to\infty$ is straightforward: $$-\int_0^1\ln\Big(\ln \frac1x\Big)x^n(1-x)^ndx=-\int_{-1/2}^{1/2}\ln\Big(\ln \frac2{2x+1}\Big)\big(\frac12+x\big)^n\big(\frac12-x\big)^ndx$$ $$\overset{x=\frac t2}{=}-\frac1{2\cdot4^n}\int_{-1}^1\ln\Big(\ln \frac2{t+1}\Big)(1-t^2)^ndt\sim-\,\frac{\ln(\ln2)}{2\cdot4^n}\sqrt\frac\pi n$$ where the Laplace' method was used in the last line.

Using the same approach we find that $$-\gamma \,B(n+1;n+1)\sim-\,\frac{\gamma}{2\cdot4^n}\sqrt\frac\pi n\,\, \text{at}\,\, n\to\infty$$ and the desired asymptotics $$\boxed{\,\,S_n\sim-\,\frac{\gamma+\ln(\ln2)}{2\cdot4^n}\sqrt\frac\pi n\,\,\, \text{at}\,\, n\to\infty\,\,}$$ The numeric check with WolframAlpha confirms the result.