Question

Clasa 11-Limite—————

Answer 1

$\lim_{n \to \infty} \frac{1+\frac{1}{2}+\frac{1}{3} +...+\frac{1}{n} }{ln (n) }= \\\[ \lim_{n \to \infty} \frac{1+\frac{1}{2}+\frac{1}{3} +...+\frac{1}{n} -ln(n)+ln(n) }{ln(n) }\]=$

$\[ \lim_{n \to \infty} (\frac{1+\frac{1}{2}+\frac{1}{3} +...+\frac{1}{n}-ln(n) }{ln(n)}+1)\]=$

$\frac{c}{ln(n)} +1 \rightarrow \boxed{1}$

$unde \ c \approx0,57 \ este \ constanta \ lui \ Euler$