Limita unui sir
Posted: Wed Sep 08, 2010 3:42 pm
Fie sirul \( (a_n)_{\small n\ge 0} \) definit prin : \( a_n=\sqrt{n^2+1}+\sqrt{n^2+2}+\ldots +\sqrt{n^2+n}-n^2-\frac n4 \) . Calculati \( \lim_{n\to\infty}\ a_n \) .
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Posted: Wed Sep 08, 2010 7:49 pm
Dac am facut bine calculele \( \frac{7}{32} \)
Posted: Wed Sep 08, 2010 11:02 pm
Eu am obţinut \( \frac{5}{24} \).
Posted: Sat Sep 11, 2010 1:35 pm
Postati si medodele sa vedem cine a gresit si cine nu.
Posted: Sat Sep 11, 2010 4:23 pm
Am folosit
\( 1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{1+x}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16} \), pentru \( x>0. \)
\( 1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{1+x}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16} \), pentru \( x>0. \)
Posted: Sun Sep 12, 2010 5:00 am
Eu am folosit ,,cleste''
\( a_n=\sum {\frac{k-\frac{n}{2}-\frac{1}{16}}{\sqrt{n^2+k}+n+\frac{1}{4}} \)
\( a_n=\sum {\frac{k-\frac{n}{2}-\frac{1}{16}}{\sqrt{n^2+k}+n+\frac{1}{4}} \)