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.