Limes ukliještenog niza

Odredite limes niza čiji je opći član

$\displaystyle a_n=\frac{\displaystyle\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+... ...{\sqrt[3]{n^3+1}}+\frac{1}{\sqrt[3]{n^3+2}}+\cdots +\frac{1}{\sqrt[3]{n^3+n}}}.$

Rješenje. Vrijedi

$\displaystyle \frac{n}{\sqrt{n^2+n}}\leq \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}}\leq \frac{n}{\sqrt{n^2+1}},$

(6.7)

$\displaystyle \frac{n}{\sqrt[3]{n^3+n}}\leq \frac{1}{\sqrt[3]{n^3+1}}+\frac{1}{\sqrt[3]{n^3+2}}+\cdots +\frac{1}{\sqrt[3]{n^3+n}}\leq \frac{n}{\sqrt[3]{n^3+1}}.$

(6.8)

Sada iz (6.7) i (6.8) slijedi

$\displaystyle \frac{\displaystyle\frac{n}{\sqrt{n^2+n}}}{\displaystyle\frac{n}{... ...ac{\displaystyle\frac{n}{\sqrt{n^2+1}}}{\displaystyle\frac{n}{\sqrt[3]{n^3+n}}}$

što nakon skraćivanja daje

$\displaystyle \frac{\sqrt[3]{n^3+1}}{\sqrt{n^2+n}}\leq a_n\leq \frac{\sqrt[3]{n^3+n}}{\sqrt{n^2+1}}.$

Izlučimo li

iz brojnika i nazivnika razlomaka s lijeve i desne strane, dobivamo

$\displaystyle \frac{n\sqrt[3]{1+\frac{1}{n^3}}}{n\sqrt{1+\frac{1}{n}}}\leq a_n\leq \frac{n\sqrt[3]{1+\frac{1}{n^2}}}{n\sqrt{1+\frac{1}{n^2}}},$

odnosno,

$\displaystyle \frac{\sqrt[3]{1+\frac{1}{n^3}}}{\sqrt{1+\frac{1}{n}}}\leq a_n \leq \frac{\sqrt[3]{1+\frac{1}{n^2}}}{\sqrt{1+\frac{1}{n^2}}}.$

Budući je

$\displaystyle \lim\limits_{n\to \infty}\frac{\sqrt[3]{1+\frac{1}{n^3}}}{\sqrt{... ...imits_{n\to \infty}\frac{\sqrt[3]{1+\frac{1}{n^2}}}{\sqrt{1+\frac{1}{n^2}}}=1,$

prema

[M1, teorem 6.7] slijedi $\lim\limits_{n\to \infty}a_n=1$ .

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