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Limes produkta     NIZOVI I REDOVI     Limes niza s produktima


Limes niza sa sumama

Izračunajte limes niza čiji je opći član:

a)
$ a_n=\displaystyle\frac{1}{n^3}\sum\limits_{k=1}^{n}k(k+1)$ ,

b)
$ a_n=\displaystyle \frac{1}{n+1}\sum_{k=1}^n (2k-1)-\frac{2n+1}{2}$ .

Rješenje.

a)
Vrijedi

$\displaystyle a_n$ $\displaystyle =\frac{1}{n^3}\sum\limits_{k=1}^{n}k(k+1)=\frac{1}{n^3}\sum\limit...
...2+k)= \frac{1}{n^3}\left(\sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}k\right)$    
  $\displaystyle =\frac{1}{n^3}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]=\frac{n(n+1)(2n+4)}{6n^3}=\frac{n^2+3n+2}{3n^2},$    

odakle slijedi

$\displaystyle \lim_{n\to \infty}a_n=\lim_{n\to \infty}\frac{n^2+3n+2}{3n^2}=\frac{1}{3}.$    

b)
Vrijedi

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

pa je

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