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