Limita unui sir

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Mateescu Constantin
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Limita unui sir

Post by Mateescu Constantin »

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 \( \lim_{n\to\infty}\ a_n \) .
Marius Mainea
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Post by Marius Mainea »

Dac am facut bine calculele \( \frac{7}{32} \) :wink:
enescu
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Post by enescu »

Eu am obţinut \( \frac{5}{24} \).
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Beniamin Bogosel
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Post by Beniamin Bogosel »

Postati si medodele sa vedem cine a gresit si cine nu.
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present. :)

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enescu
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Post by enescu »

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 \( x>0. \)
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Post by Marius Mainea »

Eu am folosit ,,cleste''

\( a_n=\sum {\frac{k-\frac{n}{2}-\frac{1}{16}}{\sqrt{n^2+k}+n+\frac{1}{4}} \)
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