<s>[INST] Solve the series $\sum\limits_{ n=\lfloor\frac{m}{2} \rfloor}^m \frac1{n}$ The floor function is confusing me a bit.  
 [/INST] Hint. m even $\implies n = \frac{m}{2} = \frac{2k}{2} = k$, m odd $\implies n = \frac{m-1}{2}$ since $\frac{m}{2} = \frac{2k+1}{2} = k + \frac{1}{2}$ for some k.
Edit:
Let $H(m) = \sum_{n=1}^{m} \frac{1}{n}$. For even m = 2k, then $\sum_{ n=\lfloor\frac{m}{2} \rfloor}^{m} \frac{1}{n} = H(2k) - H(k)$. 
Now H(2k) - H(k) is equivalent to the alternating sum = $1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2k}$, and so approaches ln(2). This can be proved using induction.
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