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.

$$\sum_{k=1}^{n}\frac{1}{2k-1}=\sum_{k=1}^{n}\frac{1}{2k}+\sum_{k=1}^{n}\frac{1}{2k(2k-1)}=\frac{H_n}{2}+\log(2)-\sum_{k>n}\frac{1}{2k(2k-1)}$$
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:

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

