# Simplifying a summation in the form $\frac{5}{i^{0.35}}$

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).

$\sum_{i=1}^{n}\frac{5}{i^{0.35}}$

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.

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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, $$\sum_{i=1}^{100} \frac{5}{i^{0.35}} \approx 148.93,\qquad \int_{1/2}^{100.5}\frac{5\;dx}{x^{0.35}}\approx 149.08 .$$
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