Đặt
$$S=\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}$$
$$S_1=\frac{1}{1^3}+\frac{1}{3^3}-\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}+\frac{1}{11^3}-...$$
$$S_2=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-...$$
$$4.\,\, Cmr: \,\, \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+...=\frac{4}{3}\left(\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-...\right)$$
Ta có
$$S_2=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+\frac{1}{5^3}-\frac{1}{6^3}+...=\left (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+.. \right )-\left ( \frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{6^3}+... \right )$$
$$=\left (\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+\frac{1}{6^3}+.. \right )-2\left ( \frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{6^3}+... \right )$$
$$=S-\frac{2}{2^3}\left ( \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+... \right )=S-\frac{2}{8}S=\frac{3}{4}S$$
$$3.\,\, Cmr:\,\, \frac{1}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^4+\left(\frac{2}{3}\right)^6+...\,\, \text{hội tụ}$$
Ta có
$$\frac{1}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^4+\left(\frac{2}{3}\right)^6+...<\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^4+\left(\frac{2}{3}\right)^6+...=\sum_{n=1}^{\infty}\left ( \frac{2}{3} \right )^n\ \Rightarrow\text{hội tụ}$$
$$2.\,\, Cmr: \,\, \frac{9}{8}<\sum_{n=1}^{\infty}\frac{1}{n^3}<\frac{5}{4}$$
Ta có đẳng thức sau:
$$\frac{9}{8}=\frac{1}{1^3}+\frac{1}{2^3}<\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+...=\frac{1}{1^3}+\sum_{n=2}^{\infty}\frac{1}{n^3}<\frac{1}{1^3}+\sum_{n=2}^{\infty}\frac{1}{n^2(n-1)}$$
$$=1+\sum_{n=2}^{\infty}\left ( \frac{1}{n(n-1)}-\frac{1}{n^2} \right )<1+\sum_{n=2}^{\infty}\left ( \frac{1}{(n-1)^2}-\frac{1}{n^2} \right )$$
$$=1+\lim_{n\to \infty}\left ( \frac{1}{2^2}-\frac{1}{n^2} \right )=\frac{5}{4}$$
$$1. \, \,S_1=\frac{1}{1^3}+\frac{1}{3^3}-\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}+\frac{1}{11^3}-...=\frac{3\pi^2\sqrt{2}}{16}$$
Xét hàm tuần hoàn $f(x)=x(\pi^2-x^2)$ với $x\in[-\pi,\pi],\, T=2\pi$
Ta dùng khai triển Fourier cho hàm trên được:
Vì hàm $f(x)$ là hàm lẻ nên $a_0=a_n=0$
$$b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx=(-1)^{n+1}\, \frac{12}{n^3}$$
$$\Rightarrow f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left ( a_n\cos(nx)+b_n\sin(nx) \right )$$
$$x(\pi^2-x^2)=12\sum_{n=1}^{\infty} (-1)^{n+1}\frac{\sin(nx)}{n^3}$$
Ta có
$$x=\frac{\pi}{2}\Rightarrow \frac{3\pi^3}{8}=12\left ( \frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-... \right )=12\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{(2n-1)^3}$$
$$\Rightarrow S_3 = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^3}=\frac{\pi^3}{32}$$
$$x=\frac{\pi}{4}\Rightarrow \frac{15\pi^3}{64.12}=\frac{1}{\sqrt{2}.1^3}-\frac{1}{2^3}+\frac{1}{\sqrt{2}.3^3}-\frac{1}{\sqrt{2}.5^3}+\frac{1}{6^3}-\frac{1}{\sqrt{2}.7^3}+....$$
$$=\left ( -\frac{1}{2^3}+\frac{1}{6^3}-\frac{1}{10^3}+... \right )+\frac{1}{\sqrt{2}}\left ( \frac{1}{1^3}+\frac{1}{3^3}-\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}+\frac{1}{11^3}-... \right )$$
$$=-\frac{1}{2^3}\left ( \frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3} \right )+\frac{S_1}{\sqrt{2}}=-\frac{S_3}{8}+\frac{S_1}{\sqrt{2}}$$
$$\Rightarrow S_1=\sqrt{2}\left ( \frac{S_3}{8}+\frac{15\pi^3}{64.12} \right )=\frac{3\pi^3\sqrt{2}}{128}$$
P.s: 
Mình chỉ là người làm thuê, ai có cách khác cứ post lên. Nhẹ tay thôi 
Bài viết đã được chỉnh sửa nội dung bởi Mrnhan: 07-04-2015 - 22:36
