演習問題

数列

${\displaystyle \sqrt{1}\left(\sqrt{1+2}-\sqrt{1}\right),}$ ${\displaystyle \sqrt{2}\left(\sqrt{2+2}-\sqrt{2}\right),}$ ${\displaystyle \sqrt{3}\left(\sqrt{3+2}-\sqrt{3}\right),\cdot \cdot \cdot ,}$ ${\displaystyle \sqrt{n}\left(\sqrt{n+2}-\sqrt{n}\right),\cdot \cdot \cdot }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\sqrt{n}\left(\sqrt{n+2}-\sqrt{n}\right)}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n}$ ${\displaystyle =\underset{n\to \infty }{lim}\sqrt{n}\left(\sqrt{n+2}-\sqrt{n}\right)}$

を求めよ．

数列

${\displaystyle \frac{1}{1},\frac{1}{2},\frac{1}{3},\cdots ,\frac{1}{n},\cdots }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{1}{n}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{1}{n}}$

を求めよ．

数列

${\displaystyle \frac{1}{1^2-1},\frac{1}{2^2-1},\frac{1}{3^2-1},\cdots \frac{1}{n^2-1}}$${\displaystyle ,\cdots }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{1}{n^2-1}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{1}{n^2-1}}$

を求めよ．

${\displaystyle \frac{{\left(-1\right)}^{1-1}}{1},\frac{{\left(-1\right)}^{2-1}}{2},\frac{{\left(-1\right)}^{3-1}}{3},\cdot \cdot \cdot ,\frac{{\left(-1\right)}^{n-1}}{n},\cdot \cdot \cdot }$ の行列，すなわち第 ${\displaystyle n}$ 項 ${\displaystyle a_n=\frac{{\left(-1\right)}^{n-1}}{n}}$ となる行列の極限

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{{\left(-1\right)}^{n-1}}{n}}$

を求めよ．

数列

${\displaystyle 1,{\left(\frac{1}{2}\right)}^2,{\left(\frac{1}{3}\right)}^3,\cdot \cdot \cdot ,{\left(\frac{1}{n}\right)}^n,\cdot \cdot \cdot }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n={\left(\frac{1}{n}\right)}^n}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}{\left(\frac{1}{n}\right)}^n}$

を求めよ．

数列

${\displaystyle \frac{5-8}{4-3},\frac{10-8}{8-3},\frac{15-8}{12-3}}$${\displaystyle ,\cdot \cdot \cdot ,\frac{5n-8}{4n-3}}$${\displaystyle ,\cdot \cdot \cdot }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{5n-8}{4n-3}}$ となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{5n-8}{4n-3}}$

を求めよ．

数列

${\displaystyle \frac{2-6}{1+8},\frac{16-6}{2+8},\frac{54-6}{3+8},\cdot \cdot \cdot }$${\displaystyle ,\frac{2n^3-6}{n+8}}$${\displaystyle ,\cdot \cdot \cdot }$

の，すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{2n^3-6}{n+8}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{2n^3-6}{n+8}}$

を求めよ．

数列

${\displaystyle \frac{4}{1+3},\frac{8}{4+3},\frac{12}{9+3},\cdot \cdot \cdot }$${\displaystyle ,\frac{4n}{n^2+3}}$${\displaystyle ,\cdot \cdot \cdot }$

の，すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{4n}{n^2+3}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{4n}{n^2+3}}$

を求めよ．

数列

${\displaystyle \frac{2+9}{5-3},\frac{2^2+9}{5^2-3},\frac{2^3+9}{5^3-3}}$${\displaystyle ,\cdot \cdot \cdot ,\frac{2^n+9}{5^n-3}}$${\displaystyle ,\cdot \cdot \cdot }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{2^n+9}{5^n-3}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{2^n+9}{5^n-3}}$

を求めよ．

数列

${\displaystyle \frac{\sqrt{3}}{\sqrt{7+4}},\frac{\sqrt{6}}{\sqrt{14+4}},\frac{\sqrt{9}}{\sqrt{21+4}},\cdot \cdot \cdot }$${\displaystyle ,\frac{\sqrt{3n}}{\sqrt{7n+4}}}$${\displaystyle ,\cdot \cdot \cdot }$

すなわち第 ${\displaystyle n}$ 項

${\displaystyle a_n=\frac{\sqrt{3n}}{\sqrt{7n+4}}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{\sqrt{3n}}{\sqrt{7n+4}}}$

を求めよ．

数列

${\displaystyle \frac{1-1}{3+1-1},\frac{1-2^3}{24+2^2-1},\frac{1-3^3}{81+3^2-1},\cdot \cdot \cdot ,\frac{1-n^3}{3n^3+n^2-1},\cdot \cdot \cdot }$

すなわち，第 ${\displaystyle n}$ 項が

${\displaystyle a_n=\frac{1-n^3}{3n^3+n^2-1}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{1-n^3}{3n^3+n^2-1}}$

を求めよ．

数列

${\displaystyle \frac{4+3-1}{2-3},\frac{32+6-1}{8-3},\frac{108+9-1}{18-3},\cdot \cdot \cdot ,\frac{4n^3+3n-1}{2n^2-3},\cdot \cdot \cdot }$

すなわち，第 ${\displaystyle n}$ 項が

${\displaystyle a_n=\frac{4n^3+3n-1}{2n^2-3}}$

となる数列の極限値

${\displaystyle \underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{4n^3+3n-1}{2n^2-3}}$

を求めよ．