Show the sequence $a_{n} = \frac{1}{n+1} + \frac{1}{n+2} ... +\frac{1}{2n}$ converges to ln 2 
Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$? 

Please help me with this homework question.
Let $\lbrace a_{n} \rbrace$ be the sequence defined by $$a_{n} = \frac{1}{n+1} + \frac{1}{n+2} ... +\frac{1}{2n}$$
for each positive integer $n$. Prove that this sequence converges to ln 2 by showing that $a_{n}$ is related to the partial sums of the series $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k.$
I know that $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k$ converges to ln 2.
I also know that
$s_{1} = 1,$
$s_{2} = 1/2,$
$s_{3} = 5/6,$
$s_{4} = 7/12,$
etc.
However, I am having a hard time understanding how $\lbrace a_{n} \rbrace$ is related to the partial sum of the given series. Any help or hints are greatly appreciated.
 A: $$\begin{align*}
1 - \frac{1}{2} &+ \frac{1}{3} - \dots -\frac{1}{2n} = \\
&=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1} - \left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}\right) \\
&=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1} - \frac{1}{2}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\
&=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1}  + \frac{1}{2}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) - \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\
&=1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n}  - \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\
&=\frac{1}{n+1} + \frac{1}{n+2} + \dots +\frac{1}{2n}\phantom{=}
\end{align*}$$
A: This sequence has popped up many times in SE.
You can write it as
$$ s_n = \sum_{k=1}^{n} \frac{1}{n+k}$$
Note that the partial sums of the series for $\log 2$ are
$$\eqalign{
  & {p_1} = 1  \cr 
  & {p_2} = 1 - \frac{1}{2} = \frac{1}{2}  \cr 
  & {p_3} = \frac{1}{2} + \frac{1}{3}  \cr 
  & {p_4} = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{1}{3} + \frac{1}{4}  \cr 
  & {p_5} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}  \cr 
  & {p_6} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6}  \cr 
  & {p_7} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}  \cr 
  & {p_8} = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \cr} $$
I guess that with Henning's answer and this, you can probably relate the $s_n$ with $p_n$.

Add: the alternative route is showing that
$$ s_n = \sum_{k=1}^{n} \frac{1}{n+k}$$
$$ s_n = \frac{1}{n}\sum_{k=1}^{n} \frac{1}{1+\frac{k}{n}}$$
Is a Riemman sum for
$$\int_0^1 \frac{dx}{x+1} = \log 2$$
A: Calculate $a_n-a_{n-1}$. It turns out to be $\frac{1}{2n-1}-\frac{1}{2n}$, that is, two successive terms in the power series for $\log 2$.
A: The answer by Aryabhata is good, but for a more general series you can do the following:
if $s_n = \sum\limits_{k = 1}^{n}f(k)$ , $t_n  = \int\limits_{1}^{n}f(x)dx$ , $d_n = s_n - t_n$
where $f$ is a positive decreasing function defined on [1,$+\infty$) such that  $\lim_{x\rightarrow\infty}f(x) = 0$ . For n = 1,2,..
then we have 
$0\leq d_k - \lim_{n\rightarrow \infty}d_n \leq f(k)$ for k = 1,2,...
You can check this directly, or you can refer Mathematical Analysis Apostol pg. 191 Thm 8.23
Let D = $ \lim_{n\rightarrow \infty}d_n$
So, $0 \leq s_n - t_n - D \leq f(n)$
Or we can write 
$s_n = t_n + D + O(f(n))$
Hence directly you get 
for $f(x) = \frac{1}{x}$
$\sum\limits_{k = 1}^{n} \frac{1}{k} = \ln(n) + D + O(\frac{1}{n})$
Using this formula you can show that 
for $p,q$ intergers s.t. $p \geq q\geq 1$
$\lim_{n\rightarrow \infty}\sum\limits_{k = qn +1}^{pn} \frac{1}{k}  = \ln(\frac{p}{q})$
for q = 1, p = 2 you get $\ln 2$
