When have enough bits of my series with non-negative terms been calculated?

Question

I have a power series with all terms non-negative which I want to evaluate to some arbitrarily set precision p (the length in binary digits of a MPFR floating-point mantissa). The result should be faithfully rounded. The issue is that I don't know when should I stop adding terms to the result variable, that is, how do I know when do I already have p + 32 accurate summed bits of the series? 32 is just an arbitrarily chosen small natural number meant to facilitate more accurate rounding to p binary digits.

This is my original series

0 <= h <= 1
series_orig(h) := sum(n = 0, +inf, a(n) * h^n)

But I actually need to calculate an arbitrary derivative of the above series (m is the order of the derivative):

series(h, m) := sum(n = m, +inf, a(n) * (n - m + 1) * ... * n * h^(n - m))

The rational number sequence a is defined like so:

a(n) := binomial(1/2, n)^2
      = (((2*n)!/(n!)) / (n! * 4^n * (2*n - 1)))^2

So how do I know when to stop summing up terms of series?

Is the following maybe a good strategy?

1. compute in p * 4 (which is assumed to be greater than p + 32).
2. at each point be able to recall the current partial sum and the previous one.
3. stop looping when the previous and current partial sums are equal if rounded to precision p + 32.
4. round to precision p and return.

Clarification

I'm doing this with MPFI, an interval arithmetic addon to MPFR. Thus the [mpfi] tag.

Attempts to get relevant formulas and equations

Guided by Eric in the comments, I have managed to derive a formula for the required working precision and an equation for the required number of terms of the series in the sum.

A problem, however, is that a nice formula for the required number of terms is not possible.

Someone more mathematically capable might instead be able to achieve a formula for a useful upper bound, but that seems quite difficult to do for all possible requested result precisions and for all possible values of m (the order of the derivative). Note that the formulas need to be easily computable so they're ready before I start computing the series.

Another problem is that it seems necessary to assume the worst case for h (h = 1) for there to be any chance of a nice formula, but this is wasteful if h is far from the worst case, that is if h is close to zero.