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?
Another approach as pointed out by achille hui in the comments: The limit can be written as a Riemann sum: $$ \lim_{n\to\infty} \sum_{i=n/2}^n \frac{1}{i} = \lim_{n\to\infty} \frac{1}{n} \sum_{i=n/2}^n \frac{1}{i/n} = \int_{1/2}^1 \frac{dx}{x} = \left.\log x\right|_{1/2}^1 = \log 2 $$
We know the famous relation: $$\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1),$$ where $\gamma$ is the Euler constant. Then we have: $$\sum_{k=n/2}^n\frac{1}{k}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^{\frac{n}{2}-1}\frac{1}{k}=\log n-\log\left(\frac{n}{2}-1\right)+o(1)=\log\left(\frac{2n}{n-2}\right)+o(1),$$ so it's clear that $$\lim_{n\to\infty}\sum_{k=n/2}^n\frac{1}{k}=\log 2.$$
Truncate the Maclaurin series for $\log(1+x)$ at the $2m$-th term, and evaluate at $x=1$. Take for example $m=10$. We get $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{19}-\frac{1}{20}.$$ Add $2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{20}\right)$, and subtract the same thing, but this time noting that $$ 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{20}\right)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}.$$ We get $$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}+\frac{1}{11}+\cdots+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}\right).$$ There is nice cancellation, and we get $$\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}.$$
$$\int_2^{n+1} \frac{1}{x-1}\,dx> \sum_2^n \frac{1}{k}>\int_2^{n+1}\frac{1}{x}\,dx$$
$$\ln n> \sum_2^n \frac{1}{k}>\ln (n+1)-\ln 2$$
$$\ln n-\ln (n/2)> \sum_{n/2}^n \frac{1}{k}>\ln (n+1)-\ln (n/2+1)$$
$$\ln 2> \sum_{n/2}^n \frac{1}{k}>\ln 2+\ln\left(\frac{n+1}{n+2}\right)\to \ln 2$$
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)
