How to prove the following limit: $\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0$? How to prove the following limit:
$$\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0?$$ 
 A: Note that 
$$u_n=\sum_{k=0}^n (-1)^k \binom{n}{k}\log(n+k)=\sum_{k=0}^n (-1)^k \binom{n}{k}\log(1+\frac{k}{n}) $$  
Now $$ \log (1+\frac{k}{n})=\int_0^{k/n}\frac{dt}{1+t}=\int_0^1\frac{kdu}{n+ku}$$
and 
$$\frac{1}{n+ku}=\int_0^1t^{n+ku-1}dt$$
Hence
$$u_n=\int\int_{[0,1]^2} (\sum_{k=0}^{n}(-1)^k\binom{n}{k}kt^{n-1+ku})dtdu$$
We have:
$$\sum_{k=0}^n (-1)^k \binom{n}{k}ky^k=-ny(1-y)^{n-1}$$
Hence:
$$u_n=-n\int\int_{[0,1]^2} t^{u+n-1}(1-t^u)^{n-1}dtdu$$ Now we integrate in $u$ first; let $x=t^u$, we have ($t\not =0, t\not =1$)
$$\int_0^1 t^u(1-t^u)^{n-1}du=\int_1^t x(1-x)^{n-1}\frac{dx}{x\log t}=-\frac{(1-t)^{n-1}}{n\log t}$$
Now we have:
$$u_n=\int_0^1t^{n-1}(1-t)^{n-1}\frac{1-t}{\log t}dt$$
And $$v_n=4^n u_n=4\int_0^1(4t(1-t))^{n-1}\frac{1-t}{\log t}dt$$
We show easily that $0\leq 4t(1-t)<1$ for $t\not =1/2$ (hence $(4t(1-t))^{n-1}\to 0$ ae) and that $\displaystyle \frac{1-t}{\log t}$ is in $L^1$; An application of the convergence dominated theorem show that $v_n\to 0$. 
A: Starting the same as Kelenner's answer, but using the Beta function, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\frac{n\,\mathrm{d}x}{n+kx}\\
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\int_0^1n\,t^{n-1+x+(k-1)x}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_0^1\int_0^1\frac nxt^{n/x}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_1^\infty\int_0^1\frac nxt^{nx}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_1^\infty\frac{n}{x+1}\frac{\Gamma(nx)\Gamma(n)}{\Gamma(n(x+1))}\,\mathrm{d}x\\
&=-\int_1^\infty\frac{n}{x+1}\mathrm{B}(nx,n)\,\mathrm{d}x\tag{1}
\end{align}
$$
We have the asymptotic
$$
\mathrm{B}(x,y)\sim\tilde{\mathrm{B}}(x,y)=\sqrt{\frac{2\pi(x+y)}{xy}}\frac{x^xy^y}{(x+y)^{x+y}}\tag{2}
$$
where for $x,y\ge n$, we have
$$
\tilde{\mathrm{B}}(x,y)\le\mathrm{B}(x,y)\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\tilde{\mathrm{B}}(x,y)\tag{3}
$$
Thus, for $x,n\ge1$,
$$
\begin{align}
\frac{n}{x+1}\mathrm{B}(nx,n)
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\frac{n}{x+1}\tilde{\mathrm{B}}(nx,n)\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{2\pi(x+1)n}{x}}\left(1+\frac1x\right)^{-nx}(x+1)^{-n-1}\\[6pt]
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}(x+1)^{-n-1}\tag{4}
\end{align}
$$
Applying $(4)$ to $(1)$ yields the upper bound
$$
\begin{align}
\hspace{-1cm}\left|\,\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)\,\right|
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\int_1^\infty(x+1)^{-n-1}\,\mathrm{d}x\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\frac1n2^{-n}\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{4\pi}n}\,4^{-n}\tag{5}
\end{align}
$$
A: Note that \begin{align}4^n[\sum_{k=0}^n(-1)^k\binom{n}{k}\ln(n+k)]&=4^n\lim_{x\rightarrow1^-}\sum_{k=0}^n(-1)^k\binom{n}{k}\ln(1-x^{n+k})\\&=4^n\lim_{x\rightarrow1^-}\sum_{k=0}^n(-1)^k\binom{n}{k}[\sum_{i=1}^\infty\frac{(x^{n+k})^i}{i}]\\&=4^n\lim_{x\rightarrow1^-}\sum_{i=1}^\infty\frac{1}{i}x^{ni}[\sum_{k=0}^n(-1)^k\binom{n}{k}(x^i)^k]\\&=4^n\lim_{x\rightarrow1^-}\sum_{i=1}^\infty\frac{1}{i}x^{ni}(1-x^i)^n\\&=\lim_{x\rightarrow1^-}\sum_{i=1}^\infty\frac{1}{i}[4x^i-4(x^i)^2]^n\\&\leq\lim_{x\rightarrow1^-}\sum_{i=1}^\infty\frac{1-x}{1-x^i}[4x^i-4(x^i)^2]^n\\&=4\lim_{x\rightarrow1^-}\sum_{i=1}^\infty\{[4x^i-4(x^i)^2]^{n-1}(x^i-x^{i+1})\}\\&=4\int_0^1(4x-4x^2)^{n-1}dx\end{align}
Since$$\lim_{n\rightarrow\infty}\int_0^1(4x-4x^2)^{n-1}dx=0$$
We have $$\lim_{n\rightarrow\infty}4^n[\sum_{k=0}^n(-1)^k\binom{n}{k}\ln(n+k)]=0.$$
