We have `1/(1-x)=(1-x)^(-1)=sum_(n=o)^oox^n` Thus `3/(2+x^3)



Click here to order this paper @Superbwriters.com. The Ultimate Custom Paper Writing Service

We have
`1/(1-x)=(1-x)^(-1)=sum_(n=o)^oox^n`
Thus
`3/(2+x^3)=3/(2(1+x^3/2))=(3/2)(1-(-x^3/2))^(-1)=sum_(n=0)^oo(-x^3/2)^n`
`=(3/2)(1-x^3/2+x^6/4-x^9/8+……..)`
Thus
`a_n=(-x^3/2)^n`
`a_(n+1)=(-x^3/2)^(n+1)`
`So`
`lim_(n->oo)|a_(n+1)/a_n|=lim_(n->oo)|-x^3/2|=|x^3/2|`
Thus series will converge if `|x^3/2|<1` `=> -2^(1/3)<x<2^(1/3)`
Thus radius of the convergence of the series is `-2^(1/3)<x<2^(1/3)`
2. `int_0^1(3x)/(1+x^3)dx=int_0^1(xsum_(n=0)^ooa_n)dx`
`=sum_(n=0)^ooint_0^1(3/2)x(-x^3/2)^ndx`
`=(3/2)sum_(n=0)^ooint_0^1(-1/2)^nx^(3n+1)dx`
`=(3/2)sum_(n=0)^oo(-1/2)^n{x^(3n+2)/(3n+2)}_0^1`
`=(3/2)sum_(n=0)^oo(-1/2)^n(1/(3n+2))`



Click here to order this paper @Superbwriters.com. The Ultimate Custom Paper Writing Service

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>