Question

Find the sum of the power series

n from 0 to ∞ ∑ {(-1)^n / (2n)! } . ( z- { pi/2} ) ^2
afterward compute the sum of the series
n from 0 to ∞ ∑ 1 / (2n)!

Answer 1

Answer on Question #50232, Math, Complex Analysis

Find the sum of the power series $n$ from 0 to $\infty \sum \left\{(-1)^{\wedge}n / (2n)!\right\} \cdot \left(z - \frac{pi}{2}\right) \wedge 2$ afterward compute the sum of the series $n$ from 0 to $\infty \sum \frac{1}{(2n)!}$

Solution:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \left(z - \frac{\pi}{2}\right)^2 = \left(z - \frac{\pi}{2}\right)^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}$$

It is known that Taylor series for $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, where $|x| < \infty$

So $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} = -\frac{1}{1!} + \frac{1}{2!} - \frac{1}{4!} + \dots = \frac{e^{j1} + e^{-j1}}{2j} = \cos(1)$, then

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(z - \frac{\pi}{2}\right)^2 = \left(z - \frac{\pi}{2}\right)^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} = \left(z - \frac{\pi}{2}\right)^2 \cos(1)$$ $$\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e^1 + e^{-1}}{2} = \cosh(1)$$

Answer: $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(z - \frac{\pi}{2}\right)^2 = \left(z - \frac{\pi}{2}\right)^2 \cos(1) \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \cosh(1)$

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