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Sa se arate ca pentru orice doua numere naturale \( k, n \) cu \( 1<k<n \) are loc relatia: \( \frac{1}{k^k}+\frac{1}{(k+1)^k}+\ldots+\frac{1}{n^k}<\frac{1}{(k-1)^k} \).

Concursul "Victor Valcovici" Braila, 2008

Imi demonstrez o inegalitate ajutatoare :

Daca \( k \leq x \leq n , x \in \mathbb{N} \) ne demonstram o inegalitate ajutatoare :
\( \frac{1}{(x-1)^{k-1}} - \frac{1}{x^{k-1}} = \frac{x^{k-1} -(x-1)^{k-1}}{(x-1)^{k-1}x^{k-1}}=\frac{x^{k-2}+x^{k-3}(x-1)+x^{k-4}(x-1)^2+ \dots +x(x-1)^{k-3} + (x-1)^{k-2}}{(x-1)^{k-1}x^{k-1}} > \frac{(k-1)(x-1)^{k-2}}{(x-1)^{k-1}x^{k-1}} > \frac{k-1}{x^k} \) .
De aici avem ca :

\( \frac{1}{x^k} < \frac{1}{k-1} \left(\frac{1}{(x-1)^{k-1}}- \frac{1}{x^{k-1}} \right) \) . Rezulta ca :

\( \sum_{x=k}^{n}{\frac{1}{x^k}} < \frac{1}{k-1} \sum_{x=k}^{n}{\left(\frac{1}{(x-1)^{k-1}}- \frac{1}{x^{k-1}} \right)} < \frac{1}{(k-1)^k} \) .
Atat