Mostrar que $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n^2\log(1+\frac{k^2}{n^2})}=\frac{{\pi}^2}{6}$

Question

¿Cómo puedo ampliar este límite siguiente? $$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n^2\log(1+\frac{k^2}{n^2})}=\frac{{\pi}^2}{6}.$$

Answer 1

Para cada $x$ en $[0,1]$, $1/(1+x)^2\leqslant1/(1+x)\leqslant 1$ por lo tanto, la integración de$0$$x$, $$ \frac{x}{1+x}\leqslant\log(1+x)\leqslant x. $$ Aplicando esto a todos los $x=k^{\color{red}{\alpha}}/n^\color{red}{\alpha}$ positivos $\color{red}{\alpha}$, uno ve que la suma $$ S_n^{(\color{red}{\alpha})}=\sum_{k=1}^n\frac1{n^\color{red}{\alpha}\log\left(1+\frac{k^{\color{red}{\alpha}}}{n^\color{red}{\alpha}}\right)} $$ es tal que $$ \sum_{k=1}^n\frac1{n^{\color{red}{\alpha}}}\frac1{\frac{k^{\color{red}{\alpha}}}{n^{\color{red}{\alpha}}}}\leqslant S_n^{(\color{red}{\alpha})}\leqslant\sum_{k=1}^n\frac1{n^\color{red}{\alpha}}\frac{1+\frac{k^\color{red}{\alpha}}{n^\color{red}{\alpha}}}{\frac{k^\color{red}{\alpha}}{n^\color{red}{\alpha}}}, $$ es decir, $$ \sum_{k=1}^n\frac1{k^\color{red}{\alpha}}\leqslant S_n^{(\color{red}{\alpha})}\leqslant\frac1{n^{\color{red}{\alpha}-1}}+\sum_{k=1}^n\frac1{k^\color{red}{\alpha}}. $$ En particular, $S_n^{(\color{red}{\alpha})}\to\infty$ al $\color{red}{\alpha}\leqslant1$ y, para cada $\color{red}{\alpha}\gt1$, $$ \lim_{n\to\infty}S_n^{(\color{red}{\alpha})}=\sum_{k=1}^\infty\frac1{k^\color{red}{\alpha}}=\zeta(\color{red}{\alpha}). $$

Answer 2

Answer 3

$$\ln (x+1)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}$$ $$f_n(x)=\frac{1}{n^2\ln(1+\frac{x^2}{n^2})}=\frac{1}{n^2(\frac{x^2}{n^2} - \frac{x^4}{2n^4} + \frac{x^4}{3n^6}+ \cdots)}=\frac{1}{x^2 - \frac{x^4}{2n^2} + \frac{x^6}{3n^4}+ \cdots}=\frac{1}{x^2}(\frac{1}{1 - \frac{x^2}{2n^2} + \frac{x^4}{3n^4}+ \cdots)}=\frac{1}{x^2}.({1 + \frac{x^2}{2n^2} - \frac{x^4}{12n^4}+ \cdots)}$$

$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n^2\ln(1+\frac{k^2}{n^2})}=\lim_{n\to\infty}\sum_{k=1}^n f_n(k)=\lim_{n\to\infty}\sum_{k=1}^n ({\frac{1}{k^2} + \frac{1}{2n^2} - \frac{k^2}{12n^4}+ \cdots)}=$$ $$=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k^2} + \lim_{n\to\infty}\frac{1}{2n^2}\sum_{k=1}^n 1 - \lim_{n\to\infty}\frac{1}{12n^4}\sum_{k=1}^n k^2+ \cdots$$

Sabemos que $$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $ a_j$ son constantes. Más información sobre suma http://en.wikipedia.org/wiki/Summation

Así podemos tener solamente un término después de límite.

$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n^2\ln(1+\frac{k^2}{n^2})}=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k^2}=\frac{{\pi}^2}{6}$$