TITLE: What do you get when you sum over the smaller half of the harmonic series?
QUESTION [8 upvotes]: More specifically, how would you evaluate the below formula?
$$\lim_{n\to\infty}\sum_{k=n/2}^{n}\frac{1}{k}$$
I know that the harmonic series starting at any point diverges, but when we limit it in this way, does the series diverge or converge? 
If it diverges:

How might you determine that?
Is there some $d$ that we can replace with $2$ to make the sequence converge?

If it converges:

What does it converge to, and how might you determine that?
The sequence must converge for any $d>2$. Is there a formula for the series generalized for any $d$?

REPLY [3 votes]: Concerning the limit itself, you already received the good answers.
What I found interesting is the problem of the partial sums for large values of $n$.
$$S_{n,d}=\sum_{k=\frac n d}^{n}\frac{1}{k}=H_n-H_{\frac{n}{d}-1}$$
Using, for large $p$, the expansion
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ apply it twice an continue with Taylor series to get
$$S_{n,d}=\log (d)+\frac{d+1}{2 n}+\frac{d^2-1}{12
   n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the respective roles of $n$ and $d$.