Suma reda rastavljanjem na parcijalne razlomke

Opći član zadanog reda možemo zapisati u obliku

$\displaystyle a_n=\frac{1}{c_n\cdot c_{n+1}},$

gdje je $\{c_n\}$ aritmetički niz s općim članom

. Rastavljanjem na parcijalne razlomke (vidi zadatak 6.10) slijedi

$\displaystyle a_n=\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right),$

pa je

-ta parcijalna suma jednaka

$\displaystyle s_k$	$\displaystyle =\sum_{n=1}^k\frac{1}{3}\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right)= \frac{1}{3}\sum_{n=1}^k\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right)$
	$\displaystyle =\frac{1}{3}\left[\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+ \cdots +\left(\frac{1}{3k-5}-\frac{1}{3k-2}\right) \right.$
	$\displaystyle \quad +\left. \left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)\right]$
	$\displaystyle =\frac{1}{3}\left(1-\frac{1}{3k+1}\right).$

Prema

[M1, definicija 6.9], suma reda je

$\displaystyle s=\lim_{k\to\infty} s_k=\lim_{k\to\infty}\frac{1}{3}\left(1-\frac{1}{3k+1}\right)=\frac{1}{3}.$

Opći član zadanog reda jednak je

$\displaystyle a_n$	$\displaystyle =\frac{n!-(n+2)!}{(n+3)!}\,(n+1)+\frac{n+1}{n+3} =\frac{n!-(n+2)(n+1)n!}{(n+3)(n+2)(n+1)n!}(\,n+1)+\frac{n+1}{n+3}$
	$\displaystyle =\left[\frac{1}{(n+1)(n+2)(n+3)}-\frac{1}{n+3}\right](n+1)+\frac{n+1}{n+3}$
	$\displaystyle =\frac{1}{(n+2)(n+3)}-\frac{n+1}{n+3}+\frac{n+1}{n+3}=\frac{1}{(n+2)(n+3)}.$

Rastavljanjem dobivenog izraza na parcijalne razlomke (vidi zadatak 6.10) slijedi

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

Sada

-tu parcijalnu sumu reda možemo zapisati u obliku

$\displaystyle s_k=\sum_{n=1}^k\left(\frac{1}{n+2}-\frac{1}{n+3}\right)= \left(\... ...rac{1}{4}-\frac{1}{5}\right)+ \cdots +\left(\frac{1}{k+2}-\frac{1}{k+3}\right),$

pa ukidanjem i primjenom

[M1, definicija 6.9], slijedi da je suma reda jednaka

$\displaystyle s=\lim_{k\to \infty}s_k=\lim_{k\to \infty}\left(\frac{1}{3}-\frac{1}{k+3}\right)=\frac{1}{3}.$