Solve the series $\sum\limits_{ n=\lfloor\frac{m}{2} \rfloor}^m \frac1{n}$ The floor function is confusing me a bit.  
 A: Hint: 
If $n=2k$ is even
$$\sum_{ n=\lfloor\frac{m}{2} \rfloor}^{m} \frac{1}{n}=\sum_{ n=k}^{2k} \frac{1}{n}=\sum_{ j=0}^{k} \frac{1}{k+j}=\frac{1}{k}\sum_{ j=0}^{k} \frac{1}{1+\frac{j}{k}}$$
which is a Riemann sum. Can you calculate the corresponding integral?
The case $n$ is odd is either similar, or you can use the fact that you only get an extra term which converges to $0$...
A: Denote your expression by $f_m$, that is
$$
f_m = \sum\limits_{n = \left\lfloor {m/2} \right\rfloor }^m {\frac{1}{n}} .
$$
If $m$ is even, then
$$
\sum\limits_{n = m/2}^m {\frac{1}{n}} ,
$$
and, as you can easily verify,
$$
f_{m+1} = f_m + \frac{1}{{m + 1}}.
$$
Hence the case $m$ odd follows immediately from the case $m$ even: $f_3 = f_2 + 1/3$, $f_5 = f_4 + 1/5$, $f_7 = f_6 + 1/7$, etc.
A: 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.
