<s>[INST] 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.
 [/INST] 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
$$\int_0^1 \frac{dx}{x+1} = \log 2$$
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