show that $\sum_{n=1}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k+1}\binom{n}{k}\log(k+1)=1$

Question

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

I tried by several ways but failed -_-

the expression is another form to e number by take exponential of both side

Answer 1

I'm pretty sure this formula is actually incorrect: I would expect the value of the sum to be the Euler-Mascheroni constant $\gamma$.

Explanation: write $\log{(k+1)}= \left. -\frac{d}{ds}\right|_{s=0} (1+k)^{-s}$. Then we have the integral representation $$ \frac{1}{(1+k)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} e^{-(1+k)x} \, dx, $$ and swapping the order of summation and integration, we have $$\begin{align*} \sum_{k=0}^{n} &(-1)^{k+1}\binom{n}{k} \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} e^{-(1+k)x} \, dx \\ &= \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} \left( \sum_{k=0}^{n} (-1)^{k+1}\binom{n}{k} e^{-(1+k)x} \right) \, dx \\ &= \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} \left( -e^{-x}(1-e^{-x})^n \right) \, dx \end{align*}$$ Doing this again for the outer sum (and noting the $n=0$ term is zero in the sum we want, so we throw it in for ease of calculation), we have $$\begin{align*} \sum_{n=0}^{\infty} & \frac{1}{n+1} \frac{1}{\Gamma(s)} \int_0^{\infty} x^{s-1} \left( -e^{-x}(1-e^{-x})^n \right) \, dx \\ &= -\frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1}e^{-x} \left(\frac{1}{1-e^{-x}}\sum_{n=0}^{\infty} \frac{1}{n+1}(1-e^{-x})^{n+1} \right) \, dx \\ &= -\frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1}e^{-x} \left(\frac{1}{1-e^{-x}}\left(-\log{(1-(1-e^{-x}))} \right)\right) \, dx \\ &= -\frac{1}{\Gamma(s)}\int_0^{\infty} \frac{x^s}{e^{x}-1} \, dx \\ &= -s\zeta(1+s) \end{align*}$$ Therefore we just have to find the coefficient of $s$ in the Taylor series of $s\zeta(1+s)$ at $s=0$. This expansion is well-known to be $$ 1 + \gamma s + O(s^2), $$ so the answer is $\gamma$ as I claimed.

(There is admittedly a worryingly large amount of limit interchange going on here.)

For a few other proofs, see Jonathan Sondow's 2003 paper on arXiv.

Answer 2

By Frullani's theorem we have: $$-\log(k+1)=\int_{0}^{+\infty}(e^{-kx}-1)\frac{dx}{x e^x} $$ hence: $$ \sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}\log(k+1)=\int_{0}^{+\infty}(1-e^{-x})^n\frac{dx}{x e^x}$$ as well as: $$ \sum_{n\geq 1}\frac{1}{n+1}\sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}\log(k+1)=\int_{0}^{+\infty}\frac{1-e^x(1-x)}{e^x-1}\frac{dx}{x e^x}=\color{red}{\gamma}$$ that is the Euler-Mascheroni constant.