Manipulating harmonic series Given the harmonic series has the summation
$$
\sum_{k=1}^n\frac1k=\ln|n| + O(1)$$
How do we show that:
$$
\sum_{k=1}^n\frac{1}{2k-1}=\ln\left|\sqrt{n}\right| + O(1)$$
 A: Hint 1:  You could try $\displaystyle\sum_{k=1}^n {1 \over 2k}$ and then subtract from $\displaystyle\sum_{k=1}^{2n} {1 \over k}$
Hint 2: $\log\left(ab\right) =  \log\left(a\right)+\log\left(b\right)$  and $\log\left(a^b\right) = b \log\left(a\right)$ 
A: Let $H_n = \sum_{k=1}^n$. Then, for example,
$$ \sum_{k=1}^3 {1 \over 2k-1} = {1 \over 1} + {1 \over 3} + {1 \over 5} $$
and you can rewrite this as
$$ \left( {1 \over 1} + {1 \over 2} + \cdots + {1 \over 6} \right) - {1 \over 2} \left( {1 \over 1} + {1 \over 2} + {1 \over 3} \right) $$
and this is $H_6 - {1 \over 2} H_3$. In general $\sum_{k=1}^n {1 \over 2k-1} = H_{2n} - H_n/2$. If  you assume $H_n \approx \log n$ then this becomes $\log (2n) - (1/2) \log n$; but $\log (2n) = \log 2 + \log n$, so your sum is approximately ${1 \over 2} \log n + \log 2$.
Of course this isn't a proof because of the approximations. But you can easily show that $\log n \le H_n \le (\log n) + 1$, by considering integrals of which $H_n$ is the lower or upper Riemann sum; from this you can find $(\log n)/2 + c_1 < H_{2n}-{1 \over 2} H_n < (\log n)/2 + c_2$ for certain constants $c_1$ and $c_2$. 
A: See this:
$$ \begin{aligned}
\sqrt a &= e^{\frac{\ln a }{2}} \\
\ln \sqrt a &= \frac{\ln a}{2}
\end{aligned} $$
