高専の数学解説

第2章 2.10

(1)

1+2+3+⋯+nn2=n(n+1)2n2=n+12n\frac{1+2+3+\cdots+n}{n^2} = \frac{\dfrac{n(n+1)}{2}}{n^2} = \frac{n+1}{2n}

lim⁡n→∞n+12n=lim⁡n→∞1+1n2=12\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{1+\dfrac{1}{n}}{2} = \frac{1}{2}

(2)

12+22+32+⋯n2n3=n(n+1)(2n+1)6n3=(n+1)(2n+1)6n2\frac{1^2+2^2+3^2+\cdots n^2}{n^3} = \frac{\dfrac{n(n+1)(2n+1)}{6}}{n^3} = \frac{(n+1)(2n+1)}{6n^2}

lim⁡n→∞(n+1)(2n+1)6n2=lim⁡n→∞(1+1n)(2+1n)6=13\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{\left(1+\dfrac{1}{n}\right)\left(2+\dfrac{1}{n}\right)}{6} = \frac{1}{3}

(3)

11⋅3+13⋅5+15⋅7+⋯+1(2n−1)⋅(2n+1)=\frac{1}{1\cdot3} + \frac{1}{3\cdot5} + \frac{1}{5\cdot7} + \cdots + \frac{1}{(2n-1)\cdot(2n+1)} =

1(2n−1)⋅(2n+1)=12(12n−1−12n+1)より\frac{1}{(2n-1)\cdot(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \text{より}

∑n=1n1(2n−1)⋅(2n+1)=12∑n=1n(12n−1−12n+1)=12(11−13+13−15+⋯+12n−1−12n+1)\sum_{n=1}^{n} \frac{1}{(2n-1)\cdot(2n+1)} = \frac{1}{2}\sum_{n=1}^{n}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \cdots + \frac{1}{2n-1} - \frac{1}{2n+1}\right)

=12(1−12n+1)=n2n+1= \frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \frac{n}{2n+1}

lim⁡n→∞n2n+1=lim⁡n→∞12+1n=12\lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2+\dfrac{1}{n}} = \frac{1}{2}

(4)

分子はa=1a=1 r=1r=1の等比級数なので

Sn2n−1=2n−12−12n−1=2n−12n−1=2−12n−1lim⁡n→∞(2−12n−1)=2\frac{S_n}{2^{n-1}} = \frac{\dfrac{2^n-1}{2-1}}{2^{n-1}} = \frac{2^n-1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}} \qquad \lim_{n \to \infty}\left(2 - \frac{1}{2^{n-1}}\right) = 2

解説: r31bn1z