Limes niza s logaritmima

Izračunajte limes niza čiji je opći član

$\displaystyle a_n=n\left[\ln\frac{1}{2}+\sum_{k=1}^n\ln{\frac{(k+1)^2}{k(k+2)}}\right].$

Rješenje. Korištenjem svojstava logaritamske funkcije i raspisivanjem članova produkta dobivamo

$\displaystyle a_n$	$\displaystyle =n\left[\ln\frac{1}{2}+\sum_{k=1}^n\ln{\frac{(k+1)^2}{k(k+2)}}\right]= n\left[\ln\frac{1}{2}+\ln\prod_{k=1}^n{\frac{(k+1)^2}{k(k+2)}}\right]$
	$\displaystyle = n\ln\left[\frac{1}{2}\cdot\prod_{k=1}^n{\frac{(k+1)^2}{k(k+2)}}\right]$
	$\displaystyle = n\ln\left[\frac{1}{2}\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\... ...{(n-1)^2}{(n-2)n}\cdot\frac{n^2}{(n-1)(n+1)}\cdot\frac{(n+1)^2}{n(n+2)}\right].$

Skraćivanjem slijedi

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

pa je zbog svojstava logaritamske funkcije, tvrdnje

[M1, teorem 6.6 (v)] i formule (6.5)

$\displaystyle \lim_{n\to \infty}a_n$	$\displaystyle =\lim_{n\to \infty}n\ln\left(\frac{n+1}{n+2}\right)= \lim_{n\to \infty}\ln\left(\frac{n+1}{n+2}\right)^n$
	$\displaystyle = \lim_{n\to \infty}\ln{\left[{\left(1-\frac{1}{n+2}\right)^{\displaystyle -(n+2)}}\right]}^{\displaystyle -\frac{n}{n+2}}$
	$\displaystyle =\ln{\left[\lim_{n\to \infty}{\left(1-\frac{1}{n+2}\right)^{\disp... ...yle\lim_{n\to \infty}}\left(\displaystyle -\frac{n}{n+2}\right)}=\ln e^{-1}=-1.$