Hipi Zhdripi i Matematikës/1237

Pasi $\lim _{n\to \infty }{\frac {1}{\sqrt[{3}]{1+n^{2}}}}=1$ , konkludojmë se relacioni i dhënë është i saktë. S h e m b u l l i 9. - Të provohet se $\lim _{n\to \infty }\prod _{k=2}^{n}\left(1-{\frac {1}{k^{2}}}\right)={\frac {1}{2}}$ . Z g j i d h j e : Produktin e dhënë e shprehim: ${\begin{aligned}\prod _{k=2}^{n}\left(\!1\!-\!{\frac {1}{k^{2}}}\!\right)\!&\!=\!\prod _{k=2}^{n}{\frac {(k-1)(k+1)}{k^{2}}}=\\&\!=\!\left(\!{\frac {1}{2}}\!\cdot \!{\frac {3}{2}}\!\right)\!\left(\!{\frac {2}{3}}\!\cdot \!{\frac {4}{3}}\!\right)\!\cdots \!\left(\!{\frac {n\!-\!2}{n\!-\!1}}\!\cdot \!{\frac {n}{n\!-\!1\!}}\!\right)\!\left(\!{\frac {n\!-\!1}{n}}\!\cdot \!{\frac {n\!+\!1}{n}}\!\right)\!=\!\\&\!=\!{\frac {1}{2}}\!\left(\!{\frac {3}{2}}\!\cdot \!{\frac {2}{3}}\!\right)\!\left(\!{\frac {4}{3}}\!\cdot \!{\frac {3}{4}}\!\right)\!\cdots \!\left(\!{\frac {n}{n\!-\!1}}\!\cdot \!{\frac {n\!-\!1}{n}}\!\right)\!\cdot \!{\frac {n\!+\!1}{n}}\!\!=\!{\frac {1}{2}}\!\cdot \!{\frac {n\!+\!1}{n}}\!,\end{aligned}}$ prandaj del: $\lim _{n\to \infty }\prod _{k=2}^{n}\left(1-{\frac {1}{k^{2}}}\right)=\lim _{n\to \infty }{\frac {1}{2}}\cdot {\frac {n+1}{n}}={\frac {1}{2}}\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)={\frac {1}{2}}$ . S h e m b u l l i 10. - Të provohet se $\lim _{n\to \infty }{\sqrt[{n}]{n}}=1$ . Z g j i d h j e : E shkruajmë se ${\sqrt[{n}]{n}}=1+h$ , ku $h>0$ . Nga kjo barazi marrim : $n=(1+h)^{n}$ ose $n=1+nh+{\binom {n}{2}}h^{2}+{\binom {n}{3}}h^{3}+\cdots +{\binom {n}{n-1}}h^{n-1}+h^{n}$ . Siç shihet, pra: ${\frac {n}{2}}>{\binom {n}{2}}h^{2}\Rightarrow 1>{\frac {n-1}{2}}h^{2}\ \mathrm {ose} \ h<{\sqrt {\frac {2}{n-1}}}$ . Meqenëse $\lim _{n\to \infty }{\sqrt {\frac {2}{n-1}}}=0$ , konkludojmë: $\lim _{n\to \infty }{\sqrt[{n}]{n}}=\lim _{n\to \infty }(1+h)=1+\lim _{n\to \infty }h=1$ . S h e m b u l l i 11. - Të provohet se $\left(\prod _{k=1}^{n}{\frac {1}{2^{k}+1}}\right)_{n=1}$ është varg komvergjent. Z g j i d h j e : Meqenëse: $a_{n}\!=\!\sum _{k=1}^{n}{\frac {1}{2^{k}\!+\!1}}\!<\!\sum _{k=1}^{n}{\frac {1}{2^{k}}}\!=\!{\frac {1}{2^{n}}}\sum _{k=1}^{n}2^{k-1}\!=\!{\frac {1}{2^{n}}}(2^{n}\!-\!1)\!=\!1\!-\!{\frac {1}{2^{n}}}\!<\!1$ d.m.th. $(\forall n\in \mathbb {N} )a_{n}<1$ , konkludojmë se vargu i dhënë është i kufizuar. Pasi kufiza $a_{n+1}$ mund të shprehet në këtë mënyrë: $a_{n+1}\!=\!\sum _{k=1}^{n+1}{\frac {1}{2^{k}+1}}\!=\!\sum _{k=1}^{n}{\frac {1}{2^{k}+1}}\!+\!{\frac {1}{2^{n+1}+1}}\!=\!a_{n}\!+\!{\frac {1}{2^{n+1}+1}}$