6.1 Power series and functions  (Page 4/8)

If ${\displaystyle \sum _{n=1}^{\infty }{a}_n{x}^n}$ converges, then $a_n{x}^n\to 0$ as $n\to \infty .$

${\displaystyle \sum _{n=1}^{\infty }{a}_n{x}^n}$ converges at $x=0$ for any real numbers $a_n.$

Given any sequence $a_n,$ there is always some $R>0,$ possibly very small, such that ${\displaystyle \sum _{n=1}^{\infty }{a}_n{x}^n}$ converges on $\left(\text{−}R,R\right).$

If ${\displaystyle \sum _{n=1}^{\infty }{a}_n{x}^n}$ has radius of convergence $R>0$ and if $\left|b_n\right|\le \left|a_n\right|$ for all n , then the radius of convergence of ${\displaystyle \sum _{n=1}^{\infty }{b}_n{x}^n}$ is greater than or equal to R .

Suppose that ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-3\right)}^n}$ converges at $x=6.$ At which of the following points must the series also converge? Use the fact that if ${\displaystyle \sum a_n{\left(x-c\right)}^n}$ converges at x , then it converges at any point closer to c than x .

1. $x=1$
2. $x=2$
3. $x=3$
4. $x=0$
5. $x=5.99$
6. $x=0.000001$

Suppose that ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x+1\right)}^n}$ converges at $x=\mathrm{-2}.$ At which of the following points must the series also converge? Use the fact that if ${\displaystyle \sum a_n{\left(x-c\right)}^n}$ converges at x , then it converges at any point closer to c than x .

1. $x=2$
2. $x=\mathrm{-1}$
3. $x=\mathrm{-3}$
4. $x=0$
5. $x=0.99$
6. $x=0.000001$

${\displaystyle \sum _{n=0}^{\infty }{a}_n{2}^n{x}^n}$

${\displaystyle \sum _{n=0}^{\infty }\frac{a_n{x}^n}{2^n}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{a_n{\pi }^n{x}^n}{e^n}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{a_n{\left(\mathrm{-1}\right)}^n{x}^n}{{10}^n}}$

${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(\mathrm{-1}\right)}^n{x}^{2n}}$

${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(\mathrm{-4}\right)}^n{x}^{2n}}$

${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(2x\right)}^n}{n}}$

${\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^n\frac{x^n}{\sqrt{n}}}$

${\displaystyle \sum _{n=1}^{\infty }\frac{n{x}^n}{2^n}}$