The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ What is the sum of the 'second half' of the harmonic series?

$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$

More precisely, what is the limit of the above sequence of partial sums?
Rewriting the sum as
$$
\sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{n/k}
$$
allows us to identify this as a Riemann sum related to the definite integral 
$$\int_1^2\frac1x\,dx=\ln 2.$$
To see that, divide the interval $[1,2]$ to $k$ equal length subintervals, and evaluate the function $f(x)=1/x$ at the right end of each subinterval. When $k\to\infty$, the Riemann sums will then tend to the value of this definite integral.
Here is still another method to calculate the limit

$$s=\lim_{k\to \infty } \, \left(\sum _{n=k+1}^{2 k} \frac{1}{n}\right)=\log (2)$$

First of all, from the definitions, we identify the partial sum as a difference of harmonic numbers

$$s_k=\sum _{n=k+1}^{2 k} \frac{1}{n}= H_{2k}-H_k$$

Now we form the "generating" sum of the partial sums

$$g(x) = \sum_{k=1}^\infty s_k x^k$$

Using the formula for the generating function of the harmonic number

$$h_1(x) = \sum_{k=1}^\infty H_k x^k = -\frac{\log(1-x)}{1-x}$$

and

$$h_2(x) = \sum_{k=1}^\infty H_{2k} x^k=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)$$

we find

$$g(x) = h_1+h_2=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)+\frac{\log (1-x)}{1-x}$$

Assuming that the limit $s = c$ exists we would have

$$g(x) = \sum_{k=1}^\infty c x^k = \frac{c}{1-x}$$

Hence our limit is

$$s = \lim_{x \to 1 }g(x)(1-x) \, $$

which is easily calculated to be $\log(2)$.