Calculus 2 Series

Question

Find the sum of the series

∞

∑ (-1)ⁿ [(x¹⁵ⁿ)÷(n!)]

n=0

It will be a function of the variable x

Answer 1

Johnny:

The trick here is to recognize that e^u = ∑₀^∞ (uⁿ / n!)

Consider u = -x¹⁵

Then ∑₀^∞ (uⁿ / n!) = ∑₀^∞ ((-x¹⁵)ⁿ / n!) = ∑₀^∞ ((-1)ⁿ(x¹⁵)ⁿ / n!) = ∑₀^∞ (((-1)x^15*n / n!)

So we see that ∑₀^∞ (((-1)x^15*n / n!) = e^{(-x^15)}

Please let me know if you have any questions!