# Please provide a function approximating the following hypergeometric series?

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.

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 $$f(m)=1+\frac{1}{4 m^2}-\frac{3}{8 m^3}+\frac{1}{2 m^4}-\frac{25}{32 m^5}+\frac{97}{64 m^6}-\frac{217}{64 m^7}+\frac{2095}{256 m^8}+O\left(\frac{1}{m^9}\right)$$

• 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? May 13, 2019 at 21:47
• @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. May 14, 2019 at 1:55
• 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? 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 $$S=\sum _{n=2}^\infty \left(\cosh \left(\frac{x}{n}\right)-1\right)$$