Question

Sa se calculeze limita :
$\displaystyle \limit\lim_{n\to\infty}\left(\dfrac{1}{1\cdot\ln1}+\dfrac{1}{2\cdot \ln 2}+\dfrac{1}{3\cdot \ln 3}+\ldots +\dfrac{1}{n\cdot \ln n}\right)$

Answer 1

$\displaystyle Aparent~se~poate~si~fara~inegalitati~si~limite~complicate~cu \\ \\ numere~prime.~Iata~la~ce~m-am~gandit: \\ \\ Notam~cu~S_n~suma~respectiva. \\ \\ \frac{S_n}{n}= \frac{1}{2n \ln 2}+ \frac{1}{3n \ln 3}+\frac{1}{3n \ln 4}...+ \frac{1}{n^2 \ln n} < \\ \\ <\frac{1}{2^2 \ln 2}+ \frac{1}{3^2 \ln 3}+ \frac{1}{4^2 \ln 4}+...+ \frac{1}{n^2 \ln n}< \\ \\ <\frac{1}{4 \ln 2}+ \frac{1}{3^2}+ \frac{1}{4^2}+...+ \frac{1}{n^2}. ~(*)$

$\displaystyle Acum,~daca~nu~vrem~sa~folosim~limita~ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}= \frac{\pi^2}{6},~putem~\\ \\ proceda~asa:~ \sum_{k=1}^n \frac{1}{k^2}<1+ \sum_{k=2}^n \frac{1}{(k-1)k}= 2-\frac{1}{n}<2$

$\displaystyle In~fine,~folosim~ultima~inegalitate~in~(*)~si~rezulta~ca~ \frac{S_n}{n}~este \\ \\ marginit,~ceea~ce~inseamna~ca~S_n~este~nemarginit,~si~fiind \\ \\ strict~crescator,~rezulta~S_n \to \infty.$