convergence of this series $a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$?? The sequence with nth term is given as
$a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$
this sequence will converge to??
what I did is:
for this we can use sandwich theorem i.e $f(x)\leq g(x)\leq h(x)$   then $lim h(x)=lim g(x)=lim f(x)$
here we can write...
$0\leq m\leq n$    
$0+n\leq m+n\leq n+n$
$\frac{1}{n}\geq \frac{1}{m+n}\geq \frac{1}{2n}$
adding this n times we get 
$1\geq\frac{1}{m+n}\geq \frac{n}{2n}$
applying lim we get 1/2..
 ..but its wrong.. how will it converge to log2??
 A: With $h=1/n$,$$\sum_{i=n}^{2n}\frac1i=\sum_{i=n}^{2n}\frac h{ih}\to\int_{x=1}^2\frac{dx}x.$$
Alternatively, using the Harmonic numbers,
$$H_{2n}-H_{n-1}=\ln(2n)+\gamma+\frac1{4n}...-\ln(n-1)-\gamma-\frac1{2n-2}+...\to\ln2.$$
A: It is a fact that $$H_n=\log n+\gamma+O(1/n)$$
Hence $H_{2n}=\log 2 +\log n+\gamma+O(1/2n)$ and $$H_{2n}-H_n=\sum_{i=n}^{2n}i^{-1}\to \log 2$$
A: Simply note that, 
$\displaystyle \begin{align} a_n = \sum\limits_{k=n}^{2n}\frac{1}{k} = \sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n-1}\frac{1}{k} &= \sum\limits_{k=1}^{2n}\frac{1}{k} - 2\sum\limits_{k=1}^{n-1}\frac{1}{2k} \\ &= \left(\sum\limits_{k=1}^{n}\frac{1}{2k-1} + \sum\limits_{k=1}^{n}\frac{1}{2k} - 2\sum\limits_{k=1}^{n}\frac{1}{2k}\right) + \frac{1}{n} \\ &= \left(\sum\limits_{k=1}^{n}\frac{1}{2k-1} - \sum\limits_{k=1}^{n}\frac{1}{2k}\right) +\frac{1}{n} \\&= \frac{1}{n} + \sum\limits_{k=1}^{2n}\frac{(-1)^{k-1}}{k} =\frac{1}{n} + \left(1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2n}\right)\end{align}$
The later is the alternating harmonic series which converges to $\ln 2$.
As, $\displaystyle \sum\limits_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} = \sum\limits_{k=1}^{\infty} (-1)^{k-1}\int_0^1 x^{k-1}\,dx = \int_0^1 \sum\limits_{k=1}^{\infty}(-1)^{k-1}x^{k-1}\,dx = \int_0^1 \frac{1}{1+x}\,dx = \ln 2$.
