I tried to come up with a closed form for the ordinary generating function for the sequence $\{\sqrt{n}\}_0^{\infty}$ but I could not. Is there a way to derive it using the recurrence relation $$a_{n+1} = \sqrt{a_n^2+1}. $$ Because if there is, it is no obvious to me how to do so.

I noticed that the task is trivial if we use a Dirichelt series generating function, namely $\zeta(s-\frac{1}{2})$ but this seems less in interesting to me than having a closed form for the ogf or perhaps even the exponential generating function.

  • $\begingroup$ It's going to very hard if not impossible to find. $\endgroup$
    – 6005
    Apr 30, 2014 at 18:47
  • 4
    $\begingroup$ Your (homework) is to find a closed-form for $\sum\sqrt{n}z^n$? (This is the polylogarithm ${\rm Li}_{-1/2}(z)$.) $\endgroup$
    – blue
    Apr 30, 2014 at 18:48
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    $\begingroup$ It is more of a something to think about, but thank you for the input. $\endgroup$
    – Pavelshu
    Apr 30, 2014 at 19:08

2 Answers 2


It does exist, defined as $g(z) = \sum_{n \ge 0} \sqrt{n} z^n$. It is even a nice function, in that it is analytic in a region around the origin (apply your favorite test). It doesn't have a representation in terms of elementary functions, however.

  • $\begingroup$ Why it doesn't have a representation in terms of elementary functions? $\endgroup$
    – lele
    May 4, 2014 at 13:39
  • $\begingroup$ @lechuza, few functions can be written in terms of elementary functions. I can't find it here on the tablet, but there was a question recently on integrals that can't be written in terms of elementary functions $\endgroup$
    – vonbrand
    May 4, 2014 at 14:03

You could try to use the integral formulas for fractional integrals and fractional derivatives (i.e., like half-weight derivatives of the power function; see for example semi-derivatives) in the theory of fractional calculus to express a generating function for this sequence. Note that the resulting generating function will not, however, be "''nice''" in the sense of it being P-recursive.


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