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.

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 $$ f(x) = {1\over 2x-1} $$
