Cómo probar el siguiente límite: $\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0$?

Question

Cómo probar el siguiente límite:

$$\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0?$$

Answer 1

Tenga en cuenta que $$u_n=\sum_{k=0}^n (-1)^k \binom{n}{k}\log(n+k)=\sum_{k=0}^n (-1)^k \binom{n}{k}\log(1+\frac{k}{n}) $$

Ahora $$ \log (1+\frac{k}{n})=\int_0^{k/n}\frac{dt}{1+t}=\int_0^1\frac{kdu}{n+ku}$$ y

$$\frac{1}{n+ku}=\int_0^1t^{n+ku-1}dt$$ Por lo tanto $$u_n=\int\int_{[0,1]^2} (\sum_{k=0}^{n}(-1)^k\binom{n}{k}kt^{n-1+ku})dtdu$$

Tenemos:

$$\sum_{k=0}^n (-1)^k \binom{n}{k}ky^k=-ny(1-y)^{n-1}$$

Por lo tanto:

$$u_n=-n\int\int_{[0,1]^2} t^{u+n-1}(1-t^u)^{n-1}dtdu$$ Now we integrate in $u$ first; let $x=t^u$, we have ($t\no =0, t\=1$)

$$\int_0^1 t^u(1-t^u)^{n-1}du=\int_1^t x(1-x)^{n-1}\frac{dx}{x\log t}=-\frac{(1-t)^{n-1}}{n\log t}$$ Ahora tenemos:

$$u_n=\int_0^1t^{n-1}(1-t)^{n-1}\frac{1-t}{\log t}dt$$

Y $$v_n=4^n u_n=4\int_0^1(4t(1-t))^{n-1}\frac{1-t}{\log t}dt$$

Se demuestra fácilmente que el $0\leq 4t(1-t)<1$ $t\not =1/2$ (por lo tanto, $(4t(1-t))^{n-1}\to 0$ ae) y que $\displaystyle \frac{1-t}{\log t}$$L^1$; Una aplicación de la convergencia dominada teorema demuestra que $v_n\to 0$.

Answer 2

A partir de la misma como Kelenner la respuesta, pero el uso de la función Beta, obtenemos $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k) &=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\frac{n\,\mathrm{d}x}{n+kx}\\ &=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\int_0^1n\,t^{n-1+x+(k-1)x}\,\mathrm{d}t\,\mathrm{d}x\\ &=-\int_0^1\int_0^1\frac nxt^{n/x}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\ &=-\int_1^\infty\int_0^1\frac nxt^{nx}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\ &=-\int_1^\infty\frac{n}{x+1}\frac{\Gamma(nx)\Gamma(n)}{\Gamma(n(x+1))}\,\mathrm{d}x\\ &=-\int_1^\infty\frac{n}{x+1}\mathrm{B}(nx,n)\,\mathrm{d}x\tag{1} \end{align} $$ Tenemos el asintótica $$ \mathrm{B}(x,y)\sim\tilde{\mathrm{B}}(x,y)=\sqrt{\frac{2\pi(x+y)}{xy}}\frac{x^xy^y}{(x+y)^{x+y}}\tag{2} $$ donde por $x,y\ge n$, tenemos $$ \tilde{\mathrm{B}}(x,y)\le\mathrm{B}(x,y)\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\tilde{\mathrm{B}}(x,y)\tag{3} $$ Así, por $x,n\ge1$, $$ \begin{align} \frac{n}{x+1}\mathrm{B}(nx,n) &\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\frac{n}{x+1}\tilde{\mathrm{B}}(nx,n)\\ &=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{2\pi(x+1)n}{x}}\left(1+\frac1x\right)^{-nx}(x+1)^{-n-1}\\[6pt] &\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}(x+1)^{-n-1}\tag{4} \end{align} $$ La aplicación de $(4)$ $(1)$se obtiene el límite superior $$ \begin{align} \hspace{-1cm}\left|\,\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)\,\right| &\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\int_1^\infty(x+1)^{-n-1}\,\mathrm{d}x\\ &=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\frac1n2^{-n}\\ &=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{4\pi}n}\,4^{-n}\tag{5} \end{align} $$

Answer 3

Tenga en cuenta que \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} Desde$$\lim_{n\rightarrow\infty}\int_0^1(4x-4x^2)^{n-1}dx=0$$ Tenemos $$\lim_{n\rightarrow\infty}4^n[\sum_{k=0}^n(-1)^k\binom{n}{k}\ln(n+k)]=0.$$