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