MHB Series Convergence: Showing Convergence & Sum Equivalence

[mathgirl1]

a) Show that sum_(n=0)^infinity (2^n x^n)/((1+x^2)^n) converges for all x in R\{-1,1}

b) Even though this is not a power series show that sum above = 1 + sum_(n=1)^infinity (2nx^n) for all -1<x<=1.

For part a by the ratio and root test we get |(2x)/(1+x^2)| but this does not have an n in it anymore to take the limit as n -> infinity. So I guess this means that the limit as n-> infinity is just (2x)/(1+x^2) which is a real number for all x. So does this mean that it converges for all x since the limit exists and is real? But this logic would mean that it would also converge for 1 and -1 since 2(1)/(1+1^2) = 2/2=1 and the same for -1. But this is cleary not true.

For part b it is clear that the term is 1 when n=0 so I understand how it would be 1 + sum_(n=1)^infinity (2^n x^n)/((1+x^2)^n) but not sure how to deduce that its the sum of 2nx^n.

Any help is much appreciated! Thanks in advance!

 

[chisigma]

a) Show that sum_(n=0)^infinity (2^n x^n)/((1+x^2)^n) converges for all x in R\{-1,1}

b) Even though this is not a power series show that sum above = 1 + sum_(n=1)^infinity (2nx^n) for all -1<x<=1.

For part a by the ratio and root test we get |(2x)/(1+x^2)| but this does not have an n in it anymore to take the limit as n -> infinity. So I guess this means that the limit as n-> infinity is just (2x)/(1+x^2) which is a real number for all x. So does this mean that it converges for all x since the limit exists and is real? But this logic would mean that it would also converge for 1 and -1 since 2(1)/(1+1^2) = 2/2=1 and the same for -1. But this is cleary not true.

For part b it is clear that the term is 1 when n=0 so I understand how it would be 1 + sum_(n=1)^infinity (2^n x^n)/((1+x^2)^n) but not sure how to deduce that its the sum of 2nx^n.

Any help is much appreciated! Thanks in advance!

Wellcome on MHB mathgirl!...

For the point a) You can set $\displaystyle z= \frac{2\ x}{1 + x^{2}}$ and the series becomes $\displaystyle S(z) = \sum_{n=0}^{\infty} z^{n}$, a geometric series which converges for $\displaystyle |z|<1 \implies |x| < 1$...

Kind regards

$\chi$ $\sigma$

 

[mathgirl1]

I am not understanding. So (2x)/(1+x^2) < 1 => 2x < 1 + x^2 => 0 < x^2 - 2x + 1 => 0 < (x-1)^2 => 0 < x -1 => 1 < x. So I can understand how the series converges for x > 1 but what about for x < 1 like the problem states?

 

[I like Serena]

Hi mathgirl! Welcome to MHB! :)

I am not understanding. So (2x)/(1+x^2) < 1 => 2x < 1 + x^2 => 0 < x^2 - 2x + 1 => 0 < (x-1)^2 => 0 < x -1 => 1 < x. So I can understand how the series converges for x > 1 but what about for x < 1 like the problem states?

I'm afraid your reasoning is not entirely correct.

You have $0 < (x-1)^2$.
Note that a square is always positive or zero.
In this case the square is zero if $x=1$.
So the inequality is satisfied if $x\ne 1$.

Furthermore, you have the extra condition that $${2x \over 1+x^2} > -1$$.
That should give you another condition.

 

[chisigma]

a) Show that sum_(n=0)^infinity (2^n x^n)/((1+x^2)^n) converges for all x in R\{-1,1}

b) Even though this is not a power series show that sum above = 1 + sum_(n=1)^infinity (2nx^n) for all -1<x<=1.

For part a by the ratio and root test we get |(2x)/(1+x^2)| but this does not have an n in it anymore to take the limit as n -> infinity. So I guess this means that the limit as n-> infinity is just (2x)/(1+x^2) which is a real number for all x. So does this mean that it converges for all x since the limit exists and is real? But this logic would mean that it would also converge for 1 and -1 since 2(1)/(1+1^2) = 2/2=1 and the same for -1. But this is cleary not true.

For part b it is clear that the term is 1 when n=0 so I understand how it would be 1 + sum_(n=1)^infinity (2^n x^n)/((1+x^2)^n) but not sure how to deduce that its the sum of 2nx^n.

Any help is much appreciated! Thanks in advance!

For part b) You can consider that... $\displaystyle 1 + \sum_{n=1}^{\infty} 2\ n\ x^{n} = 1 + 2\ x\ \frac{d}{dx} \sum_{n=0}^{\infty} x^{n} = 1 + \frac{2\ x}{(1-x)^{2}}\ (1)$

... where the series in (1) is geometric so that it converges for |x|<1...

Kind regards

$\chi$ $\sigma$

 

[mathgirl1]

Well I feel like an idiot now for part a and was making that way more difficult than it needed to be. Thank you both for the help.

If you could help me more on part b that would be great but not necessary. I guess I have had enough help at this point but anyway here goes...

So if n= 0 the sum becomes 1 + sum_{n=1}^inf (2^n x^n) / (1+x^2)^n. We can use the fact that it converges for all x in the interval -1<x<1 and so use the sum of the geometric series to get this is = 1 + 1/(1-r) = 1 + 1/(1-((2x)/(1+x^2))) = 1 + (1+x^2)/((x-1)^2) But now where?

I tried to work backwards like you suggested and I guess I do not understand the algebra cause I get that 1 + sum_{n=1}^inf (2nx^n) = 1 + 2 d/dx sum_{n=1}^inf x^n. Even if I understand the last equality you gave me I do not see how (2x)/(1-x)^2 relates to the give sum in part a.

More help is appreciated. Also, is there a tutorial on how to use math symbols and the programming that this site has to offer so i can better type this stuff up like you?

 

[chisigma]

Well I feel like an idiot now for part a and was making that way more difficult than it needed to be. Thank you both for the help.

If you could help me more on part b that would be great but not necessary. I guess I have had enough help at this point but anyway here goes...

So if n= 0 the sum becomes 1 + sum_{n=1}^inf (2^n x^n) / (1+x^2)^n. We can use the fact that it converges for all x in the interval -1<x<1 and so use the sum of the geometric series to get this is = 1 + 1/(1-r) = 1 + 1/(1-((2x)/(1+x^2))) = 1 + (1+x^2)/((x-1)^2) But now where?

I tried to work backwards like you suggested and I guess I do not understand the algebra cause I get that 1 + sum_{n=1}^inf (2nx^n) = 1 + 2 d/dx sum_{n=1}^inf x^n. Even if I understand the last equality you gave me I do not see how (2x)/(1-x)^2 relates to the give sum in part a.

More help is appreciated. Also, is there a tutorial on how to use math symbols and the programming that this site has to offer so i can better type this stuff up like you?

I apologize for the hurry I had and I intend to repair with a more detailed explanation. I suppose You know the concept of power series and derivative. A power series in the complex plane is written as $\displaystyle f(z) = \sum_{n=0}^{\infty} a_{n}\ z^{n}$ and You can demonstrate that it converges inside a circle where is $\displaystyle |z-z_{0}|< r$. Another property of a power series is that the derivative $\displaystyle f^{\ '} (z) = \sum_{n=1}^{\infty} n\ a_{n}\ z^{n-1}$ has the same circle of convergence of f(z) and vice-versa. The 'simplest' power series is the geometric series $\displaystyle f(z) = \sum_{n=0}^{\infty} z^{n} = \frac{1}{1 - z}$ for which is $z_{0}=0$ and r=1. Its derivative is $\displaystyle f^{\ '} (z) = \sum_{n=1}^{\infty} n\ z^{n-1} = \frac{1}{(1-z)^{2}}$ and it has the same circle of convergence. At this point I hope all is more clear for You and if not please don't exitate to require more explanation because 'nobody come from nothing'... Kind regards $\chi$ $\sigma$P.S. In MHB there is an excellent tutorial of LaTex and I suggest You to read it...