第2章 2.1

(1)

lim⁡n→∞2n1−3n=lim⁡n→∞21n−3=−23収束\lim_{n \to \infty} \frac{2n}{1-3n} = \lim_{n \to \infty} \frac{2}{\dfrac{1}{n}-3} = -\frac{2}{3} \quad \text{収束}

(2)

lim⁡n→∞3n2+12n2−3=lim⁡n→∞3+1n22−3n2=32収束\lim_{n \to \infty} \frac{3n^2+1}{2n^2-3} = \lim_{n \to \infty} \frac{3+\dfrac{1}{n^2}}{2-\dfrac{3}{n^2}} = \frac{3}{2} \quad \text{収束}

(3)

lim⁡n→∞n(n+4)n(n+1)(n+3)=lim⁡n→∞1n(1+4n)(1+1n)(1+3n)=0収束\lim_{n \to \infty} \frac{n(n+4)}{n(n+1)(n+3)} = \lim_{n \to \infty} \frac{\dfrac{1}{n}\left(1+\dfrac{4}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{n}\right)} = 0 \quad \text{収束}

(4)

lim⁡n→∞1+n21+n=lim⁡n→∞1n2+11n2+1n=∞正の無限大に発散\lim_{n \to \infty} \frac{1+n^2}{1+n} = \lim_{n \to \infty} \frac{\dfrac{1}{n^2}+1}{\dfrac{1}{n^2}+\dfrac{1}{n}} = \infty \quad \text{正の無限大に発散}

(5)

lim⁡n→∞(−1)n2nn−3=lim⁡n→∞(−1)n21−3n=(−1)n2n により振動\lim_{n \to \infty} \frac{(-1)^n 2n}{n-3} = \lim_{n \to \infty} \frac{(-1)^n 2}{1-\dfrac{3}{n}} = (-1)^n 2 \quad \text{n \text{により振動}}

(6)

lim⁡n→∞(−2)n3n−1=lim⁡n→∞(−23)n1−13n=0収束\lim_{n \to \infty} \frac{(-2)^n}{3^n-1} = \lim_{n \to \infty} \frac{\left(\dfrac{-2}{3}\right)^n}{1-\dfrac{1}{3^n}} = 0 \quad \text{収束}

(7)

lim⁡n→∞n(n+1−n)=lim⁡n→∞n(n+1−n)(n+1+n)n+1+n=lim⁡n→∞nn+1+n=lim⁡n→∞11+1n+1=12\lim_{n \to \infty} \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \dfrac{1}{n}} + 1} = \frac{1}{2}

収束

(8)

lim⁡n→∞1ncos⁡nπ=0\lim_{n \to \infty} \frac{1}{n} \cos n\pi = 0

収束

解説: r31bn1z