【数III数列の極限】数列の和を使って極限を求める

数列の和の公式を使う

$\displaystyle\lim_{n\rightarrow\infty}\cfrac{1\cdot2+2\cdot5+\cdots+n(3n-1)}{1^2+2^2+3^2+\cdots+n^2}$

$\displaystyle 1\cdot2+2\cdot5+\cdots+n(3n-1)=\sum_{k=1}^n k(3k-1)$
$\displaystyle=3\sum_{k=1}^n k^2-\sum_{k=1}^n k$

よって

$\displaystyle=\cfrac{3}{6}n(n+1)(2n+1)-\cfrac{1}{2}n(n+1)$
$=\cfrac{1}{2}n(n+1)(2n+1)-\cfrac{1}{2}n(n+1)$

$=\cfrac{1}{2}n(n+1)(2n+1-1)$
$=\cfrac{1}{2}n(n+1)\cdot2n=n^2(n+1)$

与式は

$\displaystyle=\lim_{n\rightarrow\infty}\cfrac{n^2(n+1)}{\cfrac{1}{6}n(n+1)(2n+1)}$
$\displaystyle=\lim_{n\rightarrow\infty}\cfrac{n^2(n+1)\times6}{\cfrac{1}{6}n(n+1)(2n+1)\times6}$
$\displaystyle=\lim_{n\rightarrow\infty}\cfrac{6n^2(n+1)}{n(n+1)(2n+1)}$

約分して

$\displaystyle=\lim_{n\rightarrow\infty}\cfrac{6n}{2n+1}$

分母・分子を $n$ で割ると

$\displaystyle=\lim_{n\rightarrow\infty}\cfrac{6}{2+\cfrac{1}{n}}$
$=\cfrac{6}{2+0}=3$ (答え)

Sn から極限を求める

$S_n=1+2+3+\cdots+n$ のとき

$\displaystyle\lim_{n\rightarrow\infty}\sqrt{S_{n+1}}-\sqrt{S_n}$

$=\displaystyle\lim_{n\rightarrow\infty}\cfrac{(\sqrt{S_{n+1}}-\sqrt{S_n})(\sqrt{S_{n+1}}+\sqrt{S_n})}{\sqrt{S_{n+1}}+\sqrt{S_n}}$
$=\displaystyle\lim_{n\rightarrow\infty}\cfrac{S_{n+1}-S_n}{\sqrt{S_{n+1}}+\sqrt{S_n}}$

$\begin{aligned}&S_{n+1}&=&1+2+3+\cdots+n+(n+1)\\-)&S_n&=&1+2+3+\cdots+n\\\hline&S_{n+1}-S_n&=&n+1\end{aligned}$

$S_n=\cfrac{1}{2}n(n+1)=\cfrac{1}{2}(n^2+n)$
$S_{n+1}=\cfrac{1}{2}(n+1)\{(n+1)+1\}$
$=\cfrac{1}{2}(n+1)(n+2)$
$=\cfrac{1}{2}(n^2+3n+2)$

これらを式に戻していくと

$=\displaystyle\lim_{n\rightarrow\infty}\cfrac{n+1}{\sqrt{\cfrac{1}{2}(n^2+3n+2)}+\sqrt{\cfrac{1}{2}(n^2+n)}}$

$=\displaystyle\lim_{n\rightarrow\infty}\cfrac{\cfrac{n}{n}+\cfrac{1}{n}}{\sqrt{\cfrac{1}{2}\Big(\cfrac{n^2}{n^2}+\cfrac{3n}{n^2}+\cfrac{2}{n^2}\Big)}+\sqrt{\cfrac{1}{2}\Big(\cfrac{n^2}{n^2}+\cfrac{n}{n^2}\Big)}}$
$=\displaystyle\lim_{n\rightarrow\infty}\cfrac{1+\cfrac{1}{n}}{\sqrt{\cfrac{1}{2}\Big(1+\cfrac{3}{n}+\cfrac{2}{n^2}\Big)}+\sqrt{\cfrac{1}{2}\Big(1+\cfrac{1}{n}\Big)}}$
$=\cfrac{1+0}{\sqrt{\cfrac{1}{2}(1+0+0)}+\sqrt{\cfrac{1}{2}(1+0)}}$
$=\cfrac{1}{2\sqrt{\cfrac{1}{2}}}$

$=\cfrac{1}{2\times\cfrac{1}{\sqrt{2}}}=\cfrac{1}{2\times\cfrac{\sqrt{2}}{2}}$
$=\cfrac{1}{\sqrt{2}}=\cfrac{\sqrt{2}}{2}$ (答え)