O limita pentru sirul b_{n}=\sum 1/sqrt{n^2+k} la puterea n
Moderators: Bogdan Posa, Laurian Filip, Beniamin Bogosel, Radu Titiu, Marius Dragoi
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
O limita pentru sirul b_{n}=\sum 1/sqrt{n^2+k} la puterea n
Sa se determine limita sirului \( b_n=\left(\sum_{k=1}^n\frac {1}{\sqrt {n^2+k}}\right)^n \)
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Avem sirul \( a_{n}=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}} \). Se observa usor ca \( \frac{n}{\sqrt{n^2+n}}\leq a_{n}\leq\frac{n}{\sqrt{n^2+1}} \). Deci \( \lim_{n\to\infty} a_{n}=1 \). Astfel, avem o nedeterminare de tipul \( 1^{\infty} \) pentru calculul limitei \( \lim_{n\to\infty}(a_{n})^{n} \). Avem:
\( \lim_{n\to\infty}(a_{n})^{n} \)=\( \lim_{n\to\infty}\left [1+(a_{n}-1)\right]^{n}=\lim_{n\to\infty}\left{\left [1+(a_{n}-1)\right]^{\frac{1}{a_{n}-1}}\right}^{n(a_{n}-1)}= e^{{\lim_{\n\to\infty}n(a_{n}-1)} \).
Acum, calculam \( \lim_{n\to\infty} n(a_{n}-1) \). Avem
\( \lim_{n\to\infty} n(a_{n}-1)=\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+1}}-1\right)+\left(\frac{n}{\sqrt{n^2+2}}-1\right)+ \ldots +\left(\frac{n}{\sqrt{n^2+n}}-1\right)=-\sum_{k=1}^{n}\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)} \).
Pe de alta parte, din inegalitati evidente avem ca
\( \frac{k}{\sqrt{n^2+n}(\sqrt{n^2+n}+n)}\leq\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)}\leq\frac{k}{\sqrt{n^2+1}(\sqrt{n^2+1}+n)} \).
Insumand cu \( k=1,2,3, \ldots , n \), vom obtine ca
\( \frac{n(n+1)}{2\{\sqrt{n^2+n}(\sqrt{n^2+n}+n)}\leq\sum_{k=1}^{n}\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)}\leq\frac{n(n+1)}{2\{\sqrt{n^2+1}(\sqrt{n^2+1}+n)} \). Cum sirul nostru este incadrat de doua siruri care tind la \( \frac{1}{4} \), rezulta ca sirul din mijloc tinde si el la \( \frac{1}{4} \), conform cu criteriul clestelui.
Prin urmare, limita noastra este egala cu \( e^{-\frac{1}{4} \).
\( \lim_{n\to\infty}(a_{n})^{n} \)=\( \lim_{n\to\infty}\left [1+(a_{n}-1)\right]^{n}=\lim_{n\to\infty}\left{\left [1+(a_{n}-1)\right]^{\frac{1}{a_{n}-1}}\right}^{n(a_{n}-1)}= e^{{\lim_{\n\to\infty}n(a_{n}-1)} \).
Acum, calculam \( \lim_{n\to\infty} n(a_{n}-1) \). Avem
\( \lim_{n\to\infty} n(a_{n}-1)=\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+1}}-1\right)+\left(\frac{n}{\sqrt{n^2+2}}-1\right)+ \ldots +\left(\frac{n}{\sqrt{n^2+n}}-1\right)=-\sum_{k=1}^{n}\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)} \).
Pe de alta parte, din inegalitati evidente avem ca
\( \frac{k}{\sqrt{n^2+n}(\sqrt{n^2+n}+n)}\leq\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)}\leq\frac{k}{\sqrt{n^2+1}(\sqrt{n^2+1}+n)} \).
Insumand cu \( k=1,2,3, \ldots , n \), vom obtine ca
\( \frac{n(n+1)}{2\{\sqrt{n^2+n}(\sqrt{n^2+n}+n)}\leq\sum_{k=1}^{n}\frac{k}{\sqrt{n^2+k}(\sqrt{n^2+k}+n)}\leq\frac{n(n+1)}{2\{\sqrt{n^2+1}(\sqrt{n^2+1}+n)} \). Cum sirul nostru este incadrat de doua siruri care tind la \( \frac{1}{4} \), rezulta ca sirul din mijloc tinde si el la \( \frac{1}{4} \), conform cu criteriul clestelui.
Prin urmare, limita noastra este egala cu \( e^{-\frac{1}{4} \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.