(5 points) By recognizing each series below as a Taylor series evaluated at a particular value of x, find the sum of each convergent series.\ A. $4 - \frac{4^3}{3!} + \frac{4^5}{5!} - \frac{4^7}{7!} + \dots + \frac{(-1)^n 4^{2n+1}}{(2n+1)!} + \dots = \sin4$ \ B. $1 + \frac{1}{7} + (\frac{1}{7})^2 + (\frac{1}{7})^3 + (\frac{1}{7})^4 + \dots + (\frac{1}{7})^n + \dots = \sin7$
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The series is 4 - 3! + 5! - 7! + ... We can recognize this as the Taylor series for e^x evaluated at x = -1. The Taylor series for e^x is given by: e^x = 1 + x + (x^2)/2! + (x^3)/3! + ... So, when we evaluate it at x = -1, we get: e^(-1) = 1 - 1 + (1/2) - (1/6) + Show more…
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