Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Power Series and Exact Values of Numerical Series

Key Questions

How do you use a power series to find the exact value of the sum of the series $pi-pi^2/2+pi^4/(4!) -pi^6/(6!) + …$ ?

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,

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

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

$=pi-1+(-1)=pi-2$

I hope that this was helpful.

Wataru · · Sep 28 2014
How do you use a power series to find the exact value of the sum of the series $1-1/2+1/3-1/4 + …$ ?

Alternating Harmonic Series

$sum_{n=1}^infty(-1)^{n-1}/n=1-1/2+1/3-1/4+cdots=ln2$

Since

$ln(1-x)=-sum_{n=1}^infty{x^n}/n$ ,

by setting $x=-1$ ,

$ln2=-sum_{n=1}^infty{(-1)^n}/n=sum_{n=1}^infty(-1)^{n-1}/n$

Wataru · · Sep 25 2014
How do you use a power series to find the exact value of the sum of the series $1+2+4/(2!) +8/(3!) +16/(4!) + …$ ?

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.

Wataru · 1 · Oct 10 2014

Questions

Power Series

View all chapters

Prev

Next