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Limita unui sir
Posted: Wed Dec 03, 2008 2:40 pm
by turcas
Calculati:
\( \lim_{n \to \infty}({1-\frac{1}{2}+\frac{1}{3} + \dots + \frac{(-1)^{n-1}}{n}}) \).
Posted: Wed Dec 03, 2008 4:10 pm
by Laurian Filip
Notam \( E_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}-\ln n \). Se Stie ca \( \lim_{n\to\infty}E_n=c\in (0,1) \)(constanta lui Euler).
Pentru \( n=2k+1 \) :
\( S_{2k+1}=1-\frac{1}{2}+...+\frac{{(-1)}^{2k}}{2k+1}=1+\frac{1}{2}+...+\frac{1}{2k+1}-2(\frac{1}{2}({1+\frac{1}{2}+...+\frac{1}{k}) \)
\( \lim_{k\to \infty} S_k=\lim_{k\to \infty}[E_{2k+1}+\ln_{2k+1}-(E_k+\ln k)]=c-c+\lim_{k\to \infty}(\ln(2k+1)-\ln k)=\lim_{k\to \infty} (\ln(2k+1)-\ln k)=\lim_{k\to \infty}\ \ln\frac{2k+1}{k}=\ln2 \).
Pentru \( n=2k+2 \)
\( S_{2k+2}=S_{2k+1}-\frac{1}{2k+2} \)
\( \lim_{k\to\infty} S_{2k+2}=\lim_{k\to\infty}(S_{2k+1}-\frac{1}{2k+2})=\lim_{k\to\infty}S_{2k+1}=\ln 2 \).
Deoarece sirul \( S_n \) are subsirul imparelor si cel al parelor cu aceeasi limita rezulta ca \( \lim_{n\to\infty}S_n=\ln2 \).
Posted: Wed Dec 03, 2008 8:50 pm
by Virgil Nicula
Identitatea Catalan (se arata usor !) : \( \sum_{k=1}^{2n}\frac {(-1)^{k-1}}{k}=\sum_{k=1}^n\frac {1}{n+k}\ \rightarrow\ \ln 2 \) .
Posted: Wed Dec 03, 2008 10:33 pm
by turcas
Virgil Nicula wrote:Identitatea Catalan (se arata usor !) : \( \sum_{k=1}^{2n}\frac {(-1)^{k-1}}{k}=\sum_{k=1}^n\frac {1}{n+k}\ \rightarrow\ \ln 2 \) .
Eu ma chinuiam de zor sa demonstrez ultima limita. Intr-un final mi-a iesit... ca sa-mi dau seama ca limita e cunoscuta.
Totusi postez solutia ca poate mai sunt "necititi" ca si mine pe forum si le prinde bine.
\( \sum_{k=1}^n\frac {1}{n+k}\ \rightarrow\ \ln 2. \)
Solutie:
Fie
\( S_n = \sum_{k=1}^n\frac {1}{n+k} \)
Avem ca
\( \left(1+\frac{1}{n} \right)^ \nearrow e \) si
\( \left(1+\frac{1}{n} \right)^{n+1} \searrow e \).
De aici vom scoate inegalitatea dubla:
\( \frac{1}{n+k} > \ln(n+k+1)-ln(n+k) > \frac{1}{n+k+1} \).
Daca insumam dupa
\( k=\overline{1,n} \) membru cu membru obtinem:
\( S_n > \ln \left( \frac{2n+1}{n+1} \right) > S_n + \frac{1}{2n+1}-\frac{1}{n+1} \).
De aici,
\( \ln \left( \frac{2n+1}{n+1} \right) + \frac{1}{n+1} - \frac{1}{2n+1} > S_n >\ln \left( \frac{2n+1}{n+1} \right) \).
Daca trecem la limita in inegalitatea de mai sus, gasim
\( \lim_{n \to \infty}{S_n}= \ln 2. \)
Posted: Wed Dec 03, 2008 11:11 pm
by Virgil Nicula
Mai exista o demonstratie cu sume Rieman, insa in clasa a XII - a. Presupunem
ca se cunoaste Identitatea Catalan sau se arata usor : \( S_{2n}\equiv\sum_{k=1}^{2n}\frac {(-1)^{k-1}}{k}=\sum_{k=1}^n\frac {1}{n+k} \) .
Deci \( \lim_{n \to \infty}S_{2n}=\lim_{n\to\infty}\ \sum_{k=1}^n\frac {1}{n+k}=\lim_{n\to\infty}\ \frac 1n\cdot\sum_{k=1}^n\frac {1}{1+\frac kn}=\int_0^1\frac {1}{1+x}\ \mathrm{dx}=\ln 2 \) .
Posted: Thu Dec 04, 2008 12:28 am
by Radu Titiu
Virgil Nicula wrote:
Mai exista o demonstratie cu sume Rieman, insa in clasa a XII - a :
Sau la nivel de clasa a XI-a se mai poate demonstra folosind faptul ca
\( 1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}-\ln n \) converge
.
Posted: Thu Dec 04, 2008 7:19 am
by Laurian Filip
Radu Titiu wrote:Virgil Nicula wrote:
Mai exista o demonstratie cu sume Rieman, insa in clasa a XII - a :
Sau la nivel de clasa a XI-a se mai poate demonstra folosind faptul ca
\( 1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}-\ln n \) converge
.
Cred ca asta am folosit eu
.
Am notat cu c limita sirului.