Numerically evaluating giant fractions

Question

I would like to know the numerical value of a giant fraction, which I got from doing some combinatoric calculation (n choose k gets huge).

For example, say I have

Sum[Binomial[2048, k] (1/5)^k (4/5)^(2048 - k), {k, 9728/25, 10752/
  25}]

This code runs fine and gives me a huge fraction, and I know its value should be somewhere between .6 and 1. However, I cannot simply use N[] or Round[] since the value gets too large/too small as Mathematica tries to work it out and reaches the precision limits of machine numbers.

What other options do I have for finding the value of this fraction?

Answer 1

one option

res = Sum[Binomial[2048, k] (1/5)^k (4/5)^(2048 - k), {k, 9728/25, 10752/25}];
N[res, 50]

Compare to

N[res]

Note that N with no options uses MachinePrecision. From help

Which is only

But note what help says

Answer 2

Better is using general symbolic solution of series:

series = Sum[Binomial[c, k] (1/5)^k (4/5)^(c - k), {k, a, b}, Assumptions -> {a > 0, b > 0, c > 0, c > b}]
(4^(-1 - a - b + c) 5^-c (4^(1 + b)
 Binomial[c, a] Hypergeometric2F1[1, a - c, 1 + a, -(1/4)] - 
4^a Binomial[c, 1 + b] Hypergeometric2F1[1, 1 + b - c, 
 2 + b, -(1/4)]))
f = series /. {a -> 9728/25, b -> 10752/25, c -> 2048} (Short expression)
N[f, 20]
(0.75358037090014508511)