What is the rate of growth of the partial sums of the reciprocals of the odd numbers?


$\sum_{1}^{n} \frac{1}{2i-1} = \sum_{1}^{2n} \frac{1}{i} - \frac{1}{2}\sum_{1}^{n} \frac{1}{i}$, and this is approximately $\ln(2n) - \frac{1}{2}\ln(n)+\frac{1}{2} \gamma = \frac{1}{2} \ln(n) + \ln(2) + \frac{1}{2} \gamma$ for large $n$.

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  • $\begingroup$ Thank you, Barry! $\gamma$ means Euler–Mascheroni constant? $\endgroup$ – Oleksandr Bondarenko Dec 13 '10 at 19:58
  • $\begingroup$ Yes, it does indeed. $\endgroup$ – Barry Smith Dec 13 '10 at 20:01

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