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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!

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    Are the $b_n$'s positive? – 2017-02-18
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    Yes :) Actually $b_n\in (0,1/2)$. I will edit it. – 2017-02-19
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    You may want to consider [Hahn series](https://en.wikipedia.org/wiki/Hahn_series) – 2017-02-20
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    Not of the desired form but may be interesting : $$\sum_{n=1}^{\infty}2^{-n(1-s)}\operatorname{Li}_s\left(-x^{-2^{-n}}\right)= \operatorname{Li}_s\left(x^{-1}\right)-\Gamma(1-s)\ln^{s-1}x, \qquad 01$$ This identity is a generalization of [this answer](http://math.stackexchange.com/a/597209/73025). – 2017-02-20
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    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. – 2017-02-21
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    *Hint:* In order to avoid contradictions from the value range of $b_n$ with your examples, it is better to write e.g. $b_1=1$ and $b_n\in (0;0.5)$ for $n\geq 2$. – 2017-02-24
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    As a special case with $00$ so that you will get $\displaystyle \int_1^x \frac{f(t)}{t}dt=\sum\limits_{n=1}^\infty (x^{a_n}-1)$ . If you know a formula here then you get a formula for the derivation and therefore an example for your question. – 2017-02-24
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    Does $\sum_{n=0}^\infty a_n$ converge absolutely or is $\sum_n |a_n|$ just bounded, but not necessarily converges? – 2017-02-25
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    I don't think that $b_1\in(0,1/2)$, but this can easily be fixed. – 2017-02-27

2 Answers 2

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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

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

$$\sum\limits_{k=1}^\infty\frac{x^k}{k!}\zeta(k+1)(1-\frac{1}{2^k})=\sum\limits_{k=1}^\infty\frac{x^k}{k!^2}\int\limits_0^\infty\frac{t^k}{e^t+1}dt=\int\limits_0^\infty\frac{e_2(xt)-1}{e^t+1}dt$$

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 $\, e_2(x)$) .

Note:

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

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

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    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. – 2017-02-24
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    We can both exploit our civic duty and undo the downvotes. I have done so (+1) for _your_ answer :-) – 2017-02-24
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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

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    Have done it too; thanks. :-) – 2017-02-24