- #1
AdkinsJr
- 150
- 0
I'm trying to review some calc, I went through the series and sequence sections pretty rapidly since my courses were all quarter-length.
I want to find x such that the series converges and find the sum.
[tex]1+x^2+2x^3+x^4+2x^5+x^6...[/tex]
[tex]=(1+x^2)+(2x^3+x^4)+(2x^5+x^6)...(2x^{2n+1}+x^{2n})[/tex]
[tex]\Sigma_{n=0}^{\infty}\left(x^{2n}+2x^{2n+1}\right)=\Sigma_{n=0}^{\infty}\left{(x^2)^n\right+\Sigma_{n=0}^{\infty}2x(x^2)^n[/tex]
So I think that the series converges for
[tex]0< \mid x^2\mid < 1 \rightarrow 0<x<1 [/tex]
and the sum is
[tex]\frac{1}{1-x^2}+\frac{2x}{1-x^2}=\frac{1+2x}{1-x^2}[/tex]
Is this correct? I don't have solution for this one... I'm not comfortable with it.
I want to find x such that the series converges and find the sum.
[tex]1+x^2+2x^3+x^4+2x^5+x^6...[/tex]
[tex]=(1+x^2)+(2x^3+x^4)+(2x^5+x^6)...(2x^{2n+1}+x^{2n})[/tex]
[tex]\Sigma_{n=0}^{\infty}\left(x^{2n}+2x^{2n+1}\right)=\Sigma_{n=0}^{\infty}\left{(x^2)^n\right+\Sigma_{n=0}^{\infty}2x(x^2)^n[/tex]
So I think that the series converges for
[tex]0< \mid x^2\mid < 1 \rightarrow 0<x<1 [/tex]
and the sum is
[tex]\frac{1}{1-x^2}+\frac{2x}{1-x^2}=\frac{1+2x}{1-x^2}[/tex]
Is this correct? I don't have solution for this one... I'm not comfortable with it.