Evaluate the summation $\sum_{k=1}^{n}{\frac{1}{2k-1}}$ I need to find this sum $$\sum_{k=1}^{n}{\frac{1}{2k-1}}$$ by manipulating the harmonic series. 
I have been given that $$\sum_{k=1}^{n}{\frac{1}{k}} = \ln(n) + C$$ where $C$ is a constant. I have given quite a bit of thought to it but have not been able to arrive at the solution.
Any method to find this summation would really help a lot.
 A: We can write $\displaystyle \sum^{n}_{r=1}\frac{1}{2r-1} = 1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}+\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}\right]$
So $\displaystyle \sum^{n}_{r=1}\frac{1}{2r-1} = \sum^{2n}_{r=1}\frac{1}{r}-\frac{1}{2}\sum^{n}_{r=1}\frac{1}{r}$
Now Using $\displaystyle \sum^{n}_{r=1}\frac{1}{r} = \ln(n)+\bf{Constant.}$
A: $$\sum_{k=1}^{n}\frac{1}{2k-1}=\sum_{k=1}^{n}\frac{1}{2k}+\sum_{k=1}^{n}\frac{1}{2k(2k-1)}=\frac{H_n}{2}+\log(2)-\sum_{k>n}\frac{1}{2k(2k-1)}$$
hence it follows that your sum behaves like $\frac{1}{2}\log(n)+O(1)$.
Through the above line and the asymptotics for harmonic numbers, we can say even more:

$$\sum_{k=1}^{n}\frac{1}{2k-1}=\frac{1}{2}\log(n)+\left(\frac{\gamma}{2}+\log(2)\right)+O\left(\frac{1}{n}\right).$$

A: One could use the Euler Summation formular: 
$$\sum_{a < x \le b} f(x) = \int_{a}^b f(x) dx + \int_{a}^{b}f'(x)\{x\}dx+f(a)\{a\}-f(b)\{b\}  $$ with $$ f(x) = {1\over 2x-1} $$
