【高校数学Ⅲ】2-1-2 無限等比数列｜問題集

\(=0\)

\(=\infty\)

\(\displaystyle =\lim_{n\to\infty}\frac{1-(\frac{2}{5})^n}{1+(\frac{2}{5})^n}\)
\(=1\)

\(\displaystyle \lim_{n\to\infty}\left\{\left(\frac{4}{3}\right)^n-\left(\frac{2}{3}\right)^n\right\}\)
\(=\infty\)

\(\displaystyle \lim_{n\to\infty}\frac{2}{(\frac{3}{2})^n-1}\)
\(=0\)

\(\displaystyle \lim_{n\to\infty}5^n\left\{1-\left(\frac{4}{5}\right)^n\right\}\)
\(=\infty\)

\(\displaystyle \lim_{n\to\infty}\left\{\frac{1-r^n}{1+r^n}\right\}\)
\(\displaystyle =\lim_{n\to\infty}\frac{(\frac{1}{r})^n-1}{(\frac{1}{r})^n+1}\)
\(=-1\)

\(\displaystyle \lim_{n\to\infty}\left\{\frac{1-r^n}{1+r^n}\right\}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1-1}{1+1}\)
\(=0\)

\(\displaystyle \lim_{n\to\infty}\left\{\frac{1-r^n}{1+r^n}\right\}\)
\(\displaystyle =\frac{1-0}{1+0}\)
\(=1\)

\(\displaystyle \lim_{n\to\infty}\left\{\frac{1-r^n}{1+r^n}\right\}\)
\(\displaystyle =\lim_{n\to\infty}\frac{(\frac{1}{r})^n-1}{(\frac{1}{r})^n+1}\)
\(=-1\)

極限値が\(1\)のとき、\(x^2+2x=1\)なので、
\(x^2+2x-1=0\)
\(x=-1\pm\sqrt{2}\)

極限値が\(0\)のとき、\(|x^2+2x|<1\)なので、
\(-1< x^2+2x< 1\)
これを解くと、
\(-1-\sqrt{2}< x< -1,-1< x< -1+\sqrt{2}\)

この漸化式を変形すると、
\(\displaystyle a_{n+1}-\frac{3}{4}=-\frac{1}{3}\left(a_n-\frac{3}{4}\right)\)
初項:\(\displaystyle \frac{1}{4}\)、公比:\(\displaystyle -\frac{1}{3}\)の等比数列なので、
\(\displaystyle a_n-\frac{3}{4}=\frac{1}{4}\left(-\frac{1}{3}\right)^{n+1}\)
\(\displaystyle a_n=\frac{1}{4}\left(-\frac{1}{3}\right)^{n+1}+\frac{3}{4}\)
よって、極限\(\displaystyle \lim_{n\to\infty}a_n\)は、
\(\displaystyle \lim_{n\to\infty}a_n=\frac{3}{4}\)