Stipulation: Would prefer polynomial asymptotic with shrinking error term and no (Riemann) Zeta functions.

Series: $${_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4}) =\ ?$$

Put differently, it looks like: $$\sum_{k=1}^{\infty}\frac{m!}{(m+k)!(m+\frac{1}{2})_k 4^{k}}$$

Where the subscript $k$ denotes the rising factorial.

Reason: I'm trying to find the summation formula for $\sum_{k=1}^{\infty}\frac{x^{2k}(\zeta(2k)-1)}{(2k)!}$,and the above hypergeometric series arose from trying to do so. Help would be much appreciated.


2 Answers 2


For hypergeometric functions of this kind, reduction formulae are generally very difficult as well as series expansions are.

However, just computing, it seems that $$f(m)=m\, {_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4}) $$ is almost a straight line with a slope equal to $1$.

Computing with illimited precision we get

$$\, {_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4}) =1+\frac 1{4m^2}+O\left(\frac 1{m^3}\right) $$

Now, if you look here, using your parameters, considering that $\frac 14$ is "small", we should have $$\, {_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4}) =1+\frac{1}{4 m^2+6 m+2}+\frac{1}{4 \left(4 m^4+20 m^3+35 m^2+25 m+6\right)}+\cdots$$ which, expanded as a series, would give $$1+\frac{1}{4 m^2}-\frac{3}{8 m^3}+O\left(\frac{1}{m^4}\right)$$

For $m=10^6$, the above truncated series would give $$\frac{16000080000144000114000037}{16000080000140000100000024}\approx 1.0000000000002499996250004999992187515000$$ while the exact value would be $$ 1.0000000000002499996250004999992187515156$$

If you want more terms, reworking the expansion of $\, {_1F}_2(1;m+\frac{1}{2};m+1;x)$ around $x=0$, we should get $$f(m)=1+\frac{1}{4 m^2}-\frac{3}{8 m^3}+\frac{1}{2 m^4}-\frac{25}{32 m^5}+\frac{97}{64 m^6}-\frac{217}{64 m^7}+\frac{2095}{256 m^8}+O\left(\frac{1}{m^9}\right)$$

  • $\begingroup$ then $f(m)=m\, {_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4})$ is what was meant, with the $m$ as a coefficient on the RHS? $\endgroup$ May 13, 2019 at 21:47
  • $\begingroup$ @user3108815. To tell the truth, in order to get an idea, I started computing a few values of $_1F_2(.)$ and made a plot to have a rough idea. $\endgroup$ May 14, 2019 at 1:55
  • $\begingroup$ No what I mean is, do I still have to divide $f(m)$ by $m$, as you have written in line one, in order to get $ {_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4})$, or is $f(m)$ approximating ${_1F}_2(1;m+\frac{1}{2};m+1;\frac{1}{4}$ as the 2nd, 3rd, and 4th equations imply? $\endgroup$ May 14, 2019 at 12:34

This is my conjecture (with the help of Mathematica) giving a finite series polynomial to replace the standard hypergeometric series.

$$\, _1F_2\left(1;m+\frac{1}{2},m+1;\frac{x^2}{4}\right)=\sum _{k=0}^{\infty } \frac{x^{2 k}}{4^k \prod _{j=1}^k \left(j+m-\frac{1}{2}\right) (j+m)}=\frac{(2 m)! \left(-\sum _{j=0}^{m-1} \left(\frac{(2 m-2)! \,x^{2 j}}{(2 j)!}\right)+(2 m-2)! \cosh (x)\right)}{(2 m-2)!\, x^{2 m}}$$

However I have no idea how to relate this formula to your original Zeta sum, other than utilizing the above formula at $m=1$ only.

Attempting to rewrite the Zeta sum myself I get

$$S=\sum _{k=1}^{\infty } \frac{ (\zeta (2 k)-1)\,x^{2 k}}{(2 k)!}$$

$$S=\sum _{k=1}^{\infty } \frac{ \zeta (2 k)\,x^{2 k}}{(2 k)!}-\sum _{k=1}^{\infty } \frac{ x^{2 k}}{(2 k)!}$$

$$S=\sum _{k=1}^{\infty } \frac{x^{2 k} } {(2 k)!}\sum _{n=1}^{\infty } \frac{1}{n^{2 k}}-\sum _{k=1}^{\infty } \frac{ x^{2 k}}{(2 k)!}$$

and on changing the order of the convergent double sum we have

$$S=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty } \frac{x^{2 k} } {(2 k)!} \frac{1}{n^{2 k}}-\sum _{k=1}^{\infty } \frac{ x^{2 k}}{(2 k)!}$$

which gives

$$S=\sum _{n=1}^\infty \left(\cosh \left(\frac{x}{n}\right)-1\right) -(\cosh (x)-1)$$

and finally $$S=\sum _{n=2}^\infty \left(\cosh \left(\frac{x}{n}\right)-1\right)$$


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