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Fie \( \sigma \in S_n \) si \( \alpha < 2 \). Sa se calculeze \( \lim\limits_{n \to \infty} \sum_{k=1}^n \frac{\sigma(k)}{k^{\alpha}} \).

Concursul interjudetean "Teodor Topan", Subiectul II

Pentru \( \alpha\leq 0 \) avem ca sirul tinde la infinit, in mod evident.
Vom demonstra acum si pentru \( \alpha\in(0,2) \)
Avem ca \( \frac{1}{n^{\alpha}}<\frac{1}{(n-1)^{\alpha}}<\ldots<\frac{1}{2^{\alpha}}<\frac{1}{1^{\alpha}} \) si totodata \( 1<2<\ldots<n-1<n \)
Conform inegalitatii rearanjarii avem ca:
\( \sum_{k=1}^{n}\frac{\sigma(k)}{k^\alpha}>\sum_{k=1}^{n}\frac{k}{k^{\alpha}}=\sum_{k=1}^{n}\frac{1}{k^{\alpha-1}} \)
Avem ca \( \alpha-1<1 \) si deci sirul \( x_n=\sum_{k=1}^{n}\frac{1}{k^{\alpha-1}} \) tinde la infinit (demonstratie cu teorema lui Lagrange pentru functia \( f(x)=\frac{1}{2-\alpha}\cdot x^{2-\alpha} \))
Rezulta deci ca \( \lim\limits_{n\to\infty}\sum_{k=1}^{n}\frac{\sigma(k)}{k^\alpha}=\infty \)

Avem ca \( \sum_{k=1}^{n}\frac{\sigma(k)}{k^\alpha}>\sum_{k=1}^{n}\frac{\sigma(k)}{k^2} \geq \sum_{k=1}^{n} \frac{1}{k^2} \cdot k \rightarrow \infty \).

Ultima inegalitate iese foarte rapid din inegalitatea rearanjamentelor, iar \( \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k} = \infty \) se demonstreaza cu Cesaro-Stolz, alegand \( b_n=\ln{n} \) crescator si divergent.

Pt. \( \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k}=\infty \):

\( \sum_{k=1}^n \frac{1}{k}= \sum_{k=0}^n (\frac {1}{2^k} + \frac{1}{2^k+1}+...+\frac{1}{2^k+2^k-1}) >\sum_{k=0}^n \frac {2^k}{2^{k+1}} =\frac{n+1}{2} \to \infty \)