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Please find a closed form for partial sum of a function

$$f(x)=a^{1/x}$$

I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.

No computer algebra systems I have tried so far could find a satisfactory solution. I believe that the expression can exist in the terms of incomplete Gamma function or its generalizations because indefinite integral of this function can be expressed in terms of incomplete Gamma function:

$$\int f(x) dx= x\sqrt[x]{a}-\operatorname{Ei}\left(\frac{\ln a}{x}\right)\ln a$$

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If an expression has a nice integral, there's no reason to believe the corresponding sum would have just as nice a representation, and vice-versa. –  J. M. Nov 8 '10 at 11:51
    
To illustrate J.M.'s point: think of $f(x)=1/x^2$ for example. Very simple to integrate, no nice formula for the partial sums. –  Hans Lundmark Nov 8 '10 at 13:55
    
It should be, however, certainly easy to get good upper and lower bounds on your sum. E.g., use am-gm or its refinements to get a lower-bound (or maybe just use the integral to get lower and upper bounds in the standard way) –  user1709 Nov 8 '10 at 14:55
    
@Hans Lundmark, there is a simple expression for that function, wolframalpha.com/input/… –  Anixx Nov 9 '10 at 0:38
    
And for any natural a, the function $f(x)=\frac 1{x^a}$ has the following partial sum (to x-1): $\frac{(-1)^{a-1}\psi^{(a-1)}(x)}{\Gamma(a)}+C$ –  Anixx Nov 9 '10 at 0:44
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1 Answer

Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$

We have

$$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2} + \frac{(\log a)^3}{3! k^3} + \cdots$$

and so

$$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) + \frac{(\log a)^3}{3!}\zeta_n(3) + \cdots$$

where $\zeta_n(r) = \sum_{k=1}^n 1/k^r.$ Thus $\zeta_n(0)=n$ and $\zeta_n(1)=H_n = 1 + 1/2 + 1/3 + \cdots + 1/n.$

Now $H_n= \log(n + 1/2) + \gamma + \epsilon(n),$ where $0< \epsilon(n)< 1/24n^2$ and $\gamma$ is the Euler-Mascheroni constant, and (by comparing the sum with the integral of $1/x^r$)

$$ \frac{1}{(r-1)(n+1)^{r-1}} < \sum_{k=n+1}^\infty \frac{1}{k^r} < \frac{1}{(r-1)n^{r-1}}.$$

Hence

$$ \frac{1}{(r-1)(n+1)^{r-1}} < \zeta(r) - \zeta_n(r) < \frac{1}{(r-1)n^{r-1}}$$

and so we have

$$ l(a) \le \sum_{k=1}^n a^{1/k} \le u(a) $$

where

$$l(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma \right) +\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n} \right)$$ $$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2n^2} \right) +\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3n^3} \right) + \cdots$$

and

$$u(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma + \frac{1}{24n^2} \right) +\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n+1} \right)$$ $$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2(n+1)^2} \right) +\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3(n+1)^3} \right) + \cdots.$$

Providing $a$ is not enormous and $n$ is sufficiently large, we only need a few terms for a good approximation to our sum.

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Somehow this doesn't seem as simple as OP was hoping for... –  Ross Millikan Nov 9 '10 at 14:46
    
@Ross Millikan You're right, but it's fun and gives us a handle on the sum. OP has ambitious hopes, although there's nothing wrong with that, of course :-) –  Derek Jennings Nov 9 '10 at 16:01
    
a weaker but much simpler lower bound is: $n[\prod_k a^{1/k}]^{1/n} = na^{H_n/n}]$ –  user1709 Nov 9 '10 at 20:34
    
If there is indeed a simple solution to this, I for one will be very surprised. –  J. M. Nov 10 '10 at 23:05
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