Each term in the series is of the form x^n / (n+1)!. F(x) is just the sum of each
of these terms for n >= 0 .
The derivate of each term is then just n * x^(n-1) / (n+1)!, and F'(x) is just the
sum of these derivates. Note that all but one of these terms (n=1) will have a factor
of x in there somewhere, so when x=0 they are also zero. Hence,
F'(0) = 1 * 0^(1-1) / 2! = 1/2.
Each time you take the derivate of a power of x, that power is reduced by 1.
When you take the tenth derivate of each of the terms, they will
each have their power of x reduced by 10. Hence, for n > 10 the tenth derivate
will still have some factor of x, so they will evaulate to 0 when x = 0.
Similarly, each term where n<10 will lose a power of x each time a derivative
is taken-- until eventually it has no power of x (i.e. it is just a constant), and will
then completely dissappear when the next deritivate is taken (since the derivative
of a constant is zero).
Hence, the only term you need consider for F''''''''''(0) is the term where n = 10:
10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 0^0 / (10+1)! = 1/11.
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