15: There is a factorial of odd numbers in the denominator, which clearly
points to the sine expansion. There is only one problem, the term
2k + 3 in the numerator.
The usual way of handling such a situation is to also create
2k + 3 in the denominator by integrating
x2k + 2,
which is something that can be easily created there.
We see that the procedure will involve integration, which is
usually handled more easily if you first denote the sum of the given series as
f (x) and then manipulate simultaneously both
sides of this equality.
There is an alternative way to work with the
2k + 3 in the numerator. Try to think of some
way to achieve cancelling with the factorial.
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