I have the following series expansion $$f(x) = \sum_{n=1}^\infty a_n x^{b_n},$$ where $a=\{a_n\}_{n=1}^\infty$ is such that $\sum_{n=1}^\infty |a_n| < \infty$ and $b=\{b_n\}_{n=1}^\infty$ such that $b_n\in (0,1/2)$ for all $n\geq 1$ and $\lim_n b_n = 0$.

Can I find a choice of $a$ and $b$ such that I have a closed form for $f$? For example:

$$f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n!} x^{1/n}, \mbox{ or } f(x) = \sum_{n=1}^\infty \frac{1}{n!} x^{1/n}$$ remind of the exponential but not quite. Any ideas?

Thank you for your support!

  • $\begingroup$ Are the $b_n$'s positive? $\endgroup$
    – zhw.
    Feb 18, 2017 at 20:05
  • $\begingroup$ Yes :) Actually $b_n\in (0,1/2)$. I will edit it. $\endgroup$
    – Martingalo
    Feb 19, 2017 at 0:04
  • $\begingroup$ You may want to consider Hahn series $\endgroup$
    – Andreas
    Feb 20, 2017 at 12:08
  • 1
    $\begingroup$ Not of the desired form but may be interesting : $$\sum_{n=1}^{\infty}2^{-n(1-s)}\operatorname{Li}_s\left(-x^{-2^{-n}}\right)= \operatorname{Li}_s\left(x^{-1}\right)-\Gamma(1-s)\ln^{s-1}x, \qquad 0<s<1,x>1$$ This identity is a generalization of this answer. $\endgroup$ Feb 20, 2017 at 21:10
  • $\begingroup$ Probably can't help but: You seem to contradict yourself "I have the following" and then "Can I find a choice of a_n and b_n". What do you have: a known expansion or a function you wish to estimate? Elsewhere you seem to want a monotonic b_n with an accumulation point of zero? If not, have you applied Gram-Schmidt to the x^(1/n) function sequentce. I have no idea whether it will work in this case; but it seems to me that it's the applicable process. $\endgroup$
    – rrogers
    Feb 21, 2017 at 18:16

2 Answers 2


I'm not sure if I understand what a closed formula is (sometimes it's good to say, which functions are allowed to use), but we can e.g. slowly converging series transform into fast converging series of which it is known that there is no closed form of it, thus deserve perhaps a separate definition.

Be $\enspace\displaystyle e_m(x):=\sum\limits_{k=0}^\infty\frac{x^k}{k!^m}\enspace$ for $\enspace m>0\enspace $ and with $\enspace x\in\mathbb{R}$ .

Be $\enspace\displaystyle f(x):=\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}x^{1/n}}{n}\enspace$ with $\enspace 0<x\leq 1$.

It follows that

$$f(e^x)=\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sum\limits_{k=0}^\infty\frac{x^k}{k!n^k}=\ln 2+\sum\limits_{k=1}^\infty\frac{x^k}{k!}\zeta(k+1)(1-\frac{1}{2^k})$$

which konvergies relatively fast for $\enspace x\in\mathbb{R}\enspace$ and it's also an extension for $\enspace f(x)$. $\enspace$ With


and $\enspace\displaystyle \int\limits_0^\infty\frac{1}{e^t+1}dt=\ln 2\enspace$ we get

$$f(x)=\int\limits_0^\infty\frac{e_2(t\ln x)}{e^t+1}dt$$

which is a closed formula by integral (if it's o.k. to use $\, e_2(x)$) .


Here is no big difference between the criterion $\sum\limits_{n=1}^\infty |a_n|<\infty$ and $\sum\limits_{n=1}^\infty a_n$ is convergent to get a formula. It decides if $x=1$ is included or perhaps not. If $\sum\limits_{n=1}^\infty |a_n|<\infty$ is obligatory then e.g for

$\displaystyle g(x):=\sum\limits_{n=1}^\infty\frac{x^{1/n}}{n^2}\enspace$ with $\enspace 0<x\leq 1$ and going the same way as above we get:

$$g(x)=\int\limits_0^\infty \frac{te_2'(t\ln x)}{e^t-1} dt=\int\limits_0^\infty\frac{(t-1)e^t+1}{(e^t-1)^2}\frac{e_2(t\ln x)-1}{\ln x} dt$$

  • 1
    $\begingroup$ To the person who likes to downvote so much without any explanation: It's an example how to create a formula in such a case and there is nothing wrong with it. $\endgroup$
    – user90369
    Feb 24, 2017 at 8:37
  • 2
    $\begingroup$ We can both exploit our civic duty and undo the downvotes. I have done so (+1) for your answer :-) $\endgroup$ Feb 24, 2017 at 11:23

Very much partial answer. Here is a graph of the function $$ y = f(x) = \sum_{n=1}^N \frac{1}{n!} x^{1/n} \quad ; \quad 0 < x,y < 10 $$ Convergence is quickly. The more converged (ipse est larger values of $\,N\,$) the darker the drawings. One of the few things one can say is that the function goes through the point $\,\color{red}{(x,y) = (1,e-1)}$ .

enter image description here

  • $\begingroup$ Have done it too; thanks. :-) $\endgroup$
    – user90369
    Feb 24, 2017 at 11:27

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