Probleme - unitatea de invatare "Limite de siruri"

Post Reply
Marcelina Popa
Bernoulli
Posts: 208
Joined: Wed Mar 05, 2008 3:25 pm
Location: Tulcea
Contact:

Probleme - unitatea de invatare "Limite de siruri"

Post by Marcelina Popa »

Cateva probleme cu limite de siruri (nivel "basic" si mediu): Lista nr. 1

O problema putzin mai grea:

Cercetati daca sirul cu termenul general

\( a_n=\frac{1^2-2^2+3^2-...+(-1)^nn^2}{n^3},\ n\ge 1 \)

are limita in \( \overline {\mathb R} \) si, in caz afirmativ, calculati-o.


Atentie: nu merge cu Cesaro-Stolz! De ce oare? :wink:
Theodor Munteanu
Pitagora
Posts: 98
Joined: Tue May 06, 2008 5:46 pm
Location: Sighetu Marmatiei

Subject

Post by Theodor Munteanu »

\(
\[
\begin{array}{l}
a_{2k} = \frac{{(1^2 + 2^2 + ... + (2k)^2 - 2(2^2 + 4^2 + ... + (2k)^2 )}}{{(2k)^3 }} = \\
= \frac{{(\frac{{2k(2k + 1)(4k + 1)}}{6} - 8\frac{{k(k + 1)(2k + 1)}}{6})}}{{8k^3 }} = \\
= ... = - \frac{{2k + 1}}{{8k^2 }}{\rm cu lim a}_{{\rm 2k}} = 0; \\
si{\rm calculand analog a}_{{\rm 2k + 1}} {\rm conchidem ca lim a}_{{\rm 2k + 1 = 0 }} \Rightarrow \lim {\rm a}_{\rm n} = 0; \\
\end{array}
\]
\)
La inceput a fost numarul. El este stapanul universului.
Post Reply

Return to “Clasa a 11-a”