Limit of definite sum equals $\ln(2)$ I have to show the following equality: 

$$\lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=\log(2)$$

I've been playing with it for almost an hour, mainly with the taylor expansion of $\ln(2)$. It looks very similar to what I need, but it has an alternating sign which sits in my way.
Can anyone point me in the right direction?

We have 
$$\sum_{k=1}^n \frac1k - \ln n\to \gamma$$
hence 
$$\sum_{k=\frac n2+1}^n \frac1k -( \ln n -\ln{\frac n2})\to 0$$
as $n\to\infty$. (Cf. pxchg1200's comment)
