Finding x such that 1+x^2+2x^3+x^4+2x^5+x^6... Converges

[AdkinsJr]

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.


1+x^2+2x^3+x^4+2x^5+x^6...

=(1+x^2)+(2x^3+x^4)+(2x^5+x^6)...(2x^{2n+1}+x^{2n})

\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

So I think that the series converges for

0< \mid x^2\mid < 1 \rightarrow 0<x<1

and the sum is

\frac{1}{1-x^2}+\frac{2x}{1-x^2}=\frac{1+2x}{1-x^2}

Is this correct? I don't have solution for this one... I'm not comfortable with it.

 

[mathman]

It looks almost right. In your original series there is no 2x term, but later you include it. Also the convergence is for -1 < x < 1.

 

[arildno]

Hmm..I'd rather rewrite this as:
2\sum_{n=0}^{\infty}x^{n}-\sum_{n}^{\infty}(x^{2})^{n}=\frac{2}{1-x}-\frac{1}{1-x^{2}}=\frac{2+2x}{1-x^{2}}-\frac{1}{1-x^{2}}=\frac{1+2x}{1-x^{2}}

 

[AdkinsJr]

It looks almost right. In your original series there is no 2x term, but later you include it.

The term was in the original series, I htink you mean the 2x factor was on the term. That factor came from the following algebra:

2x^{2n+1}=2xx^{2n}=2x(x^2)^n

Edit: Latex isn't working right, I typed 2xx^{2n} on the second step above...

Also the convergence is for -1 < x < 1.

That makes sense...

 

[ojs]

If you look at your original series there is no term there that has 2x (just 2x nothing else) in it. But if you look at your sum and put n = 0, the outcome would be 1 + 2x ... so there is a slight flaw, but easely fixed by subtracting 2x from the sigma notation.

 

[mathman]

The term was in the original series, I htink you mean the 2x factor was on the term. That factor came from the following algebra:

2x^{2n+1}=2xx^{2n}=2x(x^2)^n

Edit: Latex isn't working right, I typed 2xx^{2n} on the second step above...



That makes sense...

The 2x I was referring to was simply the missing term (as ojs pointed out).

 

[AdkinsJr]

The 2x I was referring to was simply the missing term (as ojs pointed out).

Oh yes, haha. I see now. I just typed the original series wrong. It is 1+2x+x^2+2x^3...The grouping should be (1+2x)+(x^2+2x^3)...

 

[tiny-tim]

= (1 + x² + x³ + …) + (x³ + x⁵ + …)