Seria armonica comparata cu o putere pozitiva.
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Virgil Nicula
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Seria armonica comparata cu o putere pozitiva.
Sa se arate ca \( \left(\forall\right)\ \alpha > 0\ \left(\exists\right)\ n_1\in\mathbb{N}^* \) astfel incat \( \left(\forall\right)\ n > n_1\ ,\ \sum_{k = 1}^n\frac 1k\ < \ n^{\alpha} \).
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Laurian Filip
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Problema este echivalenta cu
\( \lim_{n\to\infty} (n^a- \sum_{k=1}^n \frac{1}{k}) \geq 0 \).
Stim ca \( \lim_{n\to\infty} \sum_{k=1}^n=c+ \lim_{n\to\infty} \ln n \).
Relatia devine echivalenta cu
\( \lim_{n\to\infty}(n^a-\ln n) \geq c \)
\( \lim_{n\to\infty}( \ln{e^{n^a}} - \ln n) \geq c \)
\( \lim_{n\to\infty}(\ln \frac{e^{n^a}}{n}) \geq c \).
Fie \( k=\frac{n^a}{n} \). Rezulta \( k\geq 0 \).
\( \lim_{n\to\infty}(\ln \frac{e^{kn}}{n}) \geq c \)
\( \lim_{n\to\infty}(\ln \frac{(e^k)^n}{n}) \geq c, \)
care este adevarat pentru ca \( e^k> 1 \).
\( \lim_{n\to\infty} (n^a- \sum_{k=1}^n \frac{1}{k}) \geq 0 \).
Stim ca \( \lim_{n\to\infty} \sum_{k=1}^n=c+ \lim_{n\to\infty} \ln n \).
Relatia devine echivalenta cu
\( \lim_{n\to\infty}(n^a-\ln n) \geq c \)
\( \lim_{n\to\infty}( \ln{e^{n^a}} - \ln n) \geq c \)
\( \lim_{n\to\infty}(\ln \frac{e^{n^a}}{n}) \geq c \).
Fie \( k=\frac{n^a}{n} \). Rezulta \( k\geq 0 \).
\( \lim_{n\to\infty}(\ln \frac{e^{kn}}{n}) \geq c \)
\( \lim_{n\to\infty}(\ln \frac{(e^k)^n}{n}) \geq c, \)
care este adevarat pentru ca \( e^k> 1 \).