2
$\begingroup$

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

$\endgroup$
6
$\begingroup$

$\sum_{1}^{n} \frac{1}{2i-1} = \sum_{1}^{2n} \frac{1}{i} - \frac{1}{2}\sum_{1}^{n} \frac{1}{i}$, 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 $n$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.