Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My math is not incredibly strong and perhaps I have just not been searching for the right terms, but I have a summation that is part of an algorithm I've been working on and would really like to reduce it to just a formula, but am really struggling to find a solution (if one exists).


Can anyone point me in the right direction as to how to approach this, or is likely not possible to reduce down to just a formula? Thanks very much in advance.

share|cite|improve this question
Well, it's $5\cdot H_{n,7/20}$, but that's no simple formula as such. – NikolajK Feb 5 '13 at 15:51
Just to be clear, $i$ is a variable and not the imaginary unit, correct? Regards – Amzoti Feb 5 '13 at 15:54

If you are interested in how it behaves for large $n$, you could try an approximation like $$ \sum_{i=1}^n \frac{5}{i^{0.35}} \approx \int_{1/2}^{n+1/2}\frac{5\;dx}{x^{0.35}} $$ For example, $$ \sum_{i=1}^{100} \frac{5}{i^{0.35}} \approx 148.93,\qquad \int_{1/2}^{100.5}\frac{5\;dx}{x^{0.35}}\approx 149.08 . $$

share|cite|improve this answer
Yeah, so reagading evaluation of the integral, I tried it and the error for the approximation $"(50/13)\cdot 2^{7/20}\cdot ((2n+1)^{13/20}-1)-94/625 "$ starts with $4\cdot 10^{-2}$ and falls below $10^{-3}$ after $n=25$. – NikolajK Feb 5 '13 at 16:32
The error in this kind of integral approximation is often mainly due to approximation errors in the first few terms. So if we are willing to have uglier code, we can store the answers for $n=1$, $2$, and so on up to (say) $10$, and for $n\gt 10$ use the stored value for $10$ plus the integral approximation for the rest of the sum. – André Nicolas Feb 5 '13 at 16:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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