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?

$$\int_2^{n+1} \frac{1}{x-1}\,dx> \sum_2^n \frac{1}{k}>\int_2^{n+1}\frac{1}{x}\,dx$$
$$\ln n> \sum_2^n \frac{1}{k}>\ln (n+1)-\ln 2$$
$$\ln n-\ln (n/2)> \sum_{n/2}^n \frac{1}{k}>\ln (n+1)-\ln (n/2+1)$$
$$\ln 2> \sum_{n/2}^n \frac{1}{k}>\ln 2+\ln\left(\frac{n+1}{n+2}\right)\to \ln 2$$
