# Power Series and Exact Values of Numerical Series

## Key Questions

• Since

1-x^2/{2!}+{x^4}/{4!}-{x^6}/{6!}+cdots=cosx,

1-pi^2/{2!}+{pi^4}/{4!}-{pi^6}/{6!}+cdots=cos(pi)=-1

Therefore,

pi-pi^2/{2!}+{pi^4}/{4!}-{pi^6}/{6!}+cdots

=pi-1+(1-pi^2/{2!}+{pi^4}/{4!}-{pi^6}/{6!}+cdots)

$= \pi - 1 + \left(- 1\right) = \pi - 2$

I hope that this was helpful.

• Alternating Harmonic Series

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$

Since

$\ln \left(1 - x\right) = - {\sum}_{n = 1}^{\infty} \frac{{x}^{n}}{n}$,

by setting $x = - 1$,

$\ln 2 = - {\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n}}{n} = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n$

• Since

1+x+x^2/{2!}+x^3/{3!}+x^4/{4!}+cdots=e^x,

by replacing $x$ by $2$,

1+2+2^2/{2!}+2^3/{3!}+2^4/{4!}+cdots=e^2.

I hope that this was helpful.