can anyone help me find the sum of this power series?

Ian FowlerNov 16, 2019This is a power series which produces an infinite polynomial in x. It does not converge to a single value. The polynomial is:1 + (2/3)x + (4/5)x^2 + (8/7)x^3 + (16/9)x^4 + ....Using the ratio test we evaluate the limit as n ---> infinity of the ratio a(n+1)/a(n) where a(n) = 2^n*x^n/(2n+1) . This ratio simplifies to: 2x(2n+1)/(2n+3)As n ---> infinity this limit becomes 2x. Therefore this series will converge to a single value when the value of x lies in the interval 0 < x < 1/2.Other than that I am not sure what you are asking for.

andrei.frit02Nov 16, 2019Thank you for the replay. Sorry if I wasn't clear enough.I must find the function that has this series expansion.

Ian FowlerNov 16, 2019Fair point. I should have figured that out. I got this far:1/(1-y) = 1 + y + y^2 + y^3 + y^4 + ... for |y| < 1Replace y with 2x and we get:1/(1-2x) = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + ... for |x| < 1/2This gives us the numerators in the terms of the series (i. e. the numerators for a Geometric Series)I can't see a way to alter 1/(1-2x) to get the odd numbers in the denominator. generating functions maybe? I'll keep at it.

Aneesh SrinivasFeb 5, 2020;okay 1 slight correction on the radius of convergence i said on the last step,it should be for -1/2≤x<1/2 and x≠0(arctanh(√2x)/√(2x) is undefined there).

andrei.frit02Feb 10, 2020Thank you so much. After all it was quite simple, but I lack creativity, I guess. :))

This is a power series which produces an infinite polynomial in x. It does not converge to a single value. The polynomial is:

1 + (2/3)x + (4/5)x^2 + (8/7)x^3 + (16/9)x^4 + ....

Using the ratio test we evaluate the limit as n ---> infinity of the ratio a(n+1)/a(n) where

a(n) = 2^n*x^n/(2n+1) . This ratio simplifies to: 2x(2n+1)/(2n+3)

As n ---> infinity this limit becomes 2x. Therefore this series will converge to a single value when the value of x lies in the interval 0 < x < 1/2.

Other than that I am not sure what you are asking for.

Thank you for the replay. Sorry if I wasn't clear enough.I must find the function that has this series expansion.

Fair point. I should have figured that out. I got this far:

1/(1-y) = 1 + y + y^2 + y^3 + y^4 + ... for |y| < 1

Replace y with 2x and we get:

1/(1-2x) = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + ... for |x| < 1/2

This gives us the numerators in the terms of the series (i. e. the numerators for a Geometric Series)

I can't see a way to alter 1/(1-2x) to get the odd numbers in the denominator. generating functions maybe? I'll keep at it.

;okay 1 slight correction on the radius of convergence i said on the last step,

it should be for -1/2≤x<1/2 and x≠0(arctanh(√2x)/√(2x) is undefined there).

Thank you so much. After all it was quite simple, but I lack creativity, I guess. :))