Lời giải:
Câu 1: Ta có:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+...+(-1)^nx^n+...\ (x \in [-1;1])$
$\to\displaystyle\int_{0}^{x} \dfrac{1}{1+x} dx = x - \dfrac{x^2}2+...+(-1)^n \dfrac{x^{n+1}}{n+1}+...$
$\to \ln (1+x)=\sum\limits_{n=1}^{\infty}(-1)^{n-1} \dfrac{x^2}{n}\ (x\in [-1;1])$ (đpcm)
Câu 2: Áp dụng bất đẳng thức Svacxơ có:
$\left|\dfrac{(1-x^2)\sin a + 2x\cos a}{1+x^2}\right|\\=\dfrac{|(1-x^2)\sin a + 2x\cos a|}{1+x^2}\leq\dfrac{\sqrt{1+x^4-2x^2+4x^2}}{1+x^2}\\\leq \dfrac{\sqrt{(1+x^2)^2}}{1+x^2}=1$ (đpcm)
Câu 3: Ta có: $\sum\limits_{n = 1}^k \dfrac{1}{i^2n^2}=\dfrac1{i^2} \sum\limits_{n=1}^k \dfrac{1}{n^2}$
Vì $\dfrac1{i^2} \sum\limits_{n=1}^k \dfrac{1}{n^2} < 2.\dfrac1{i^2} $
$\to \sum\limits_{i=1}^{q}\ \sum\limits_{n=1}^k \dfrac1{i^2n^2} < \sum\limits_{i=1}^{q} \dfrac2{i^2} <4$ (đpcm)
