第2章 2.9

(1)

$\lim_{n\to\infty}\left\{2+\left(-\frac{1}{3}\right)^n\right\}=2+0=2 \quad \text{収束}$

(2)

$\lim_{n\to\infty}\left\{(-2)^n+\left(-\frac{1}{2}\right)^n\right\}=\pm\infty+0 \quad \text{振動}$

(3)

$\lim_{n\to\infty}\{\log_2(n^2-1)-\log_2 n^2\}=\lim_{n\to\infty}\log_2\frac{n^2-1}{n^2}=\lim_{n\to\infty}\log_2\frac{1-\dfrac{1}{n^2}}{1}=\log_2 1=0 \quad \text{収束}$

(4)

$\lim_{n\to\infty}\frac{1}{\sqrt{2n+1}-\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{n}}{(\sqrt{2n+1}-\sqrt{n})(\sqrt{2n+1}+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{n}}{2n+1-n}=\lim_{n\to\infty}\frac{\sqrt{\dfrac{2}{n}+\dfrac{1}{n^2}}+\sqrt{\dfrac{1}{n}}}{1+\dfrac{1}{n}}=0$

収束

(5)

$\lim_{n\to\infty}\frac{4n}{\sqrt{n^2+1}+n}=\lim_{n\to\infty}\frac{4}{\sqrt{1+\dfrac{1}{n^2}}+1}=\frac{4}{2}=2 \quad \text{収束}$

(6)

$\lim_{n\to\infty}(n-\sqrt{n^2+3n})=\lim_{n\to\infty}\frac{(n-\sqrt{n^2+3n})(n+\sqrt{n^2+3n})}{n+\sqrt{n^2+3n}}=\lim_{n\to\infty}\frac{n^2-(n^2+3n)}{n+\sqrt{n^2+3n}}=\lim_{n\to\infty}\frac{-3n}{n+\sqrt{n^2+3n}}$

$=\lim_{n\to\infty}\frac{-3}{1+\sqrt{1+\dfrac{3}{n}}}=-\frac{3}{2}$