Help me please with 2 questions:
Find limit of
1.$\displaystyle a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}$
2.$\displaystyle \underset{n \to \infty }{\lim} \left [ \frac{1}{n^{2}}+...+\frac{1}{(2n)^{2}} \right ]$
Thanks!
Considering that is...
$\displaystyle \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n-2} \frac{1}{k} - \ln (n-2) \} =\gamma$ (1)
$\displaystyle \lim_{n \rightarrow \infty} \{\sum_{k=1}^{2n} \frac{1}{k} - \ln 2 n \} =\gamma$ (2)
... subtracting (1) from (2) You obtain...
$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=n-1}^{2n} \frac{1}{k} = \lim_{n \rightarrow \infty} \{\ln 2 n - \ln (n-2)\}= \ln 2$ (3)
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$
You could also try another approach:
$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]$$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]$
$\displaystyle = \left[ \lim_{n \to \infty}\frac{1}{n-1}+\lim_{n \to \infty}\frac{1}{n}+ \lim_{n \to \infty}\sum_{r=1}^{n}\frac{1}{n+r}\right]$
$\displaystyle = \left[ 0+0+\lim_{n \to \infty}\sum_{r=1}^{n}\left(\frac{1}{n} \right)\frac{1}{1+{r \over n}}\right]$
$\displaystyle =\int_{0}^{1} \frac{1}{1+x} dx$
$\displaystyle =[\ln(1+x)]^1_0$
$\displaystyle =\ln(2)$
$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]=0$$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]$