Riemann sum of n = 0 to infinity, (-1)^n * x^(2n) / n! for all real x, then f''(0) = ?
I changed this to (-x^2)^n / n! in order to make it similar to the e^x maclaurin series. I then used e^(-x^2) as my function for the series and took the second derivative which I got to be -2 + 6x^2 - 5x^4 +... I plugged in 0 for x and thus the answer is -2. Is this logic correct?