A sum to the power of e e

Calculus Level pending

If

S = n = 1 1 2 n ( k = 0 1 2 k k ! ) n S = {\sum_{n=1}^{\infty}\frac{1}{2n}\left ( \sum_{k=0}^{\infty}\frac{1}{2^{k}k!} \right )^{n}}

Suppose e S e^{S} may be written as

a b c × e \frac{a}{b-\sqrt{c \times e}}

Then find a + b + c a + b + c .

OBS:

  • e e is the base of the natural logarithm.


The answer is 5.

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