Sir cu logaritmi

[Marius Mainea]

Fie sirul \( (x_n)_{n\ge 1} \) definit prin \( x_n=\sum_{k=n}^{2n-1}{\sqrt[n]{\log_k(k+1)}}. \) Sa se demonstreze ca :

a) \( x_n <x_{n+1}; \)

b) \( [n(x_n-n)]=0,\ (\forall) n \ge 3 \).

Gabriel Dospinescu rev. Arhimede /2007

[Laurian Filip]

a) \( x_{n+1} \) este egal cu o suma de n+1 termeni supraunitari, deci \( x_{n+1}>n+1 \)

Pe de alta parte, din teorema lui Bernoulli
\( (1+\frac{1}{n})^n \geq 2 \geq \log_k{k+1} \)
\( 1+\frac{1}{n} \geq \sqrt[n]{\log_k{k+1} \)

\( x_n<n \cdot (1+ \frac{1}{n})=n+1 \)

deci \( x_n<n+1<x_{n+1} \)

[Laurian Filip]

b) \( (1+\frac{1}{k})^{k}<e \)
\( k \ln(1+\frac{1}{k})<1 \)
\( \ln(k+1)-\ln(k)<\frac{1}{k} \)
\( \frac{\ln(k+1)}{\ln{k}}<1+\frac{1}{k\ln{k}} \)


\( \log_k{k+1}<1+\frac{1}{k\ln{k}}<1+\frac{1}{n}<(1+\frac{1}{n^2})^n \) (prima relatie pt ca \( k \geq n \), iar a doua din Bernoulli), de unde
\( \sqrt[n]{\log_k{k+1}}< 1+\frac{1}{n^2} \).

Rezulta ca
\( x_n <n (1+\frac{1}{n^2})=n+\frac{1}{n} \)
\( [n(x_n-n)]<[n(n+\frac{1}{n}-n)]=1 \)

In punctul a) am demonstrat ca \( x_n>n \)

Deci \( 0 \leq [n(x_n-n)] < 1 \)

\( [n(x_n-n)]=0 \).