Hipi Zhdripi i Matematikës/1236

Stampa:Hipi Zhdripi i Matematikës kryefleta Stampa:Z g j i d h j e Meqenëse: Stampa:Dygishta$1+2^{-1}+2^{-2}+\dots +2^{-n}=\frac{1-2^{-(n+1)}}{1-2^{-1}}=2[1-{\left(\frac{1}{2}\right)}^{n+1}]$ dhe

Stampa:Dygishta$1+3^{-1}+3^{-2}+\dots +3^{-n}=\frac{1-3^{-(n+1)}}{1-3^{-1}}=2[1-{\left(\frac{1}{3}\right)}^{n+1}]$

konkludojmë se:

$a_n=\frac{4}{3}\cdot \frac{1-{\left(\frac{1}{2}\right)}^{n+1}}{1-{\left(\frac{1}{3}\right)}^{n+1}}$

prej nga marrim :

$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{4}{3}\cdot \frac{1-{\left(\frac{1}{2}\right)}^{n+1}}{1-{\left(\frac{1}{3}\right)}^{n+1}}=\frac{4}{3}\cdot \frac{\underset{n\to \mathrm{\infty }}{\lim }\left[1-{\left(\frac{1}{2}\right)}^{n+1}\right]}{\underset{n\to \mathrm{\infty }}{\lim }\left[1-{\left(\frac{1}{3}\right)}^{n+1}\right]}=\frac{4}{3}$

Stampa:S h e m b u l l i Të caktohet limiti i vargut, kufiza e përgjithshme e të cilit është $a_n=\frac{\sin n^2}{7n}$.

Stampa:Z g j i d h j e Meqenëse $(\mathrm{\forall }x\in ℝ)|\sin x|⩽1$, marrim:

$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{\sin n^2}{7n}=\frac{1}{7}\underset{n\to \mathrm{\infty }}{\lim }\frac{\sin n^2}{n}=0$.

Stampa:S h e m b u l l i Të caktohet limiti i vargut, kufiza e përgjithshme e të cilit është $a_n=\frac{1+3+5+\dots +(2n-1)}{n+1}-\frac{2n+1}{2}$.

Stampa:Z g j i d h j e Pasi që

$a_n=\frac{1+3+5+\dots +(2n-1)}{n+1}-\frac{2n+1}{2}=\frac{n^2}{n+1}-\frac{2n+1}{2}=-\frac{3n+1}{2(n+1)}$,

marrim:

$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=-\underset{n\to \mathrm{\infty }}{\lim }\frac{3n+1}{2(n+1)}=-\frac{\underset{n\to \mathrm{\infty }}{\lim }(3+\frac{1}{n})}{\underset{n\to \mathrm{\infty }}{\lim }(2+\frac{2}{n})}=-\frac{3}{2}$

Stampa:S h e m b u l l i Të provohet se

$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{\sqrt[3]{n^3+1}}+\frac{1}{\sqrt[3]{n^3+2}}+\cdots +\frac{1}{\sqrt[3]{n^3+n}})=1$

Stampa:Z g j i d h j e Zgjidhjen e detyrës e bazojmë në jobarazinë e dyfishtë:

$(\mathrm{\forall }n\in \mathbb{N})\frac{n}{\sqrt[3]{n^3+1}}<\frac{1}{\sqrt[3]{n^3+1}}+\frac{1}{\sqrt[3]{n^3+2}}+\cdots +\frac{1}{\sqrt[3]{n^3+n}}<\frac{n}{\sqrt[3]{n^3}}$

prej nga marrim

$\frac{1}{\sqrt[3]{1+\frac{1}{n^2}}}<\frac{1}{\sqrt[3]{n^3+1}}+\frac{1}{\sqrt[3]{n^3+2}}+\cdots +\frac{1}{\sqrt[3]{n^3+n}}<1$