Manipulating harmonic series

Given the harmonic series has the summation

$$\sum_{k=1}^n\frac1k=\ln|n| + O(1)$$

How do we show that:

$$\sum_{k=1}^n\frac{1}{2k-1}=\ln\left|\sqrt{n}\right| + O(1)$$

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I cleared up your notation. In particular it's not true that $\sum_{k=1}^n {1 \over k} = \ln n$ exactly; it's only approximately true. –  Michael Lugo Sep 30 '11 at 0:08
Thanks - you beat me to tidying up my TeX. Hopefully I've clarified the post. –  oliland Sep 30 '11 at 0:15

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 $\log\left(a^b\right) = b \log\left(a\right)$

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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 $c_2$.

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\begin{aligned} \sqrt a &= e^{\frac{\ln a }{2}} \\ \ln \sqrt a &= \frac{\ln a}{2} \end{aligned}
That's part of Henry's answer with $b=1/2$, and it's also not helpful on its own in this situation. –  anon Feb 8 '12 at 1:07