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

A. We have

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

Using the formula for the cosine expansion in a power series $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$ is easy to see that

If $z \to \left[z - \frac{\pi}{2}\right]$, then

$$\begin{aligned} & \sum_{n=0}^{\infty} \frac{(-1)^n \left(z - \frac{\pi}{2}\right)^{2n}}{(2n)!} = \cos \left(z - \frac{\pi}{2}\right) = \frac{e^{i\left(z - \frac{\pi}{2}\right)} + e^{-i\left(z - \frac{\pi}{2}\right)}}{2} = \frac{e^{iz} e^{\frac{i\pi}{2}} + e^{-iz} e^{-\frac{i\pi}{2}}}{2} \\ & \quad = \left\{e^{\frac{i\pi}{2}} = i; e^{-\frac{i\pi}{2}} = -i\right\} = \frac{i e^{iz} - i e^{-iz}}{2} = \frac{i}{2} \left(e^{iz} - e^{-iz}\right) = \frac{i}{2} \sum_{n=0}^{\infty} \frac{(1 - (-1)^n) i^n}{n!} z^n \\ & \quad = \frac{i}{2} \sum_{n=0}^{\infty} \frac{2i^{2n+1}}{(2n+1)!} z^{2n+1} = \frac{i}{2} \sum_{n=0}^{\infty} \frac{2i(-1)^n}{(2n+1)!} z^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} z^{2n+1} = \sin z \end{aligned}$$

B. We have

$$\sum_{n=0}^{\infty} \frac{1}{(2n)!}$$

Using the formula for the hyperbolic cosine expansion in a power series

$$\sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} = \operatorname{csch} z$$

Let $z = 1$, then

$$\sum_{n=0}^{\infty} \frac{1^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \operatorname{csch} 1$$

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