## mf1021 3 years ago find the radius of convergence of the taylor series around x=0 for ln(1/(1+2x))

1. amistre64

i spose we would nee to generate the power series for it first

2. amistre64

$[ln\frac{1}{1+2x}]'=\frac{1/(1+2x)'}{1/(1+2x)}$ $[ln\frac{1}{1+2x}]'=\frac{-2/(1+2x)^2}{1/(1+2x)}$ $[ln\frac{1}{1+2x}]'=\frac{-2/(1+2x)}{1/1}\to\ -\frac{2}{1+2x}$

3. amistre64

maybe?

4. amistre64

lol -ln(1+2x) = -2/(1+2x)

5. amistre64

-2+4x-8x^2+16x^3-32x^4+64x^5 ... ----------------- 1+2x ) -2 (-2-4x) 4x (4x+8x^2) -8x^2 (-8x^2-16x^3)

6. afranklin12

2 i think

7. experimentX

somewhere i had seen something like this integration dln(1+2x)/dx and expand it as power series.

8. amistre64

soo, a relevant power series would be:$ln(\frac{1}{1+2x})=\int\sum (-1)^{n+1}\ 2^{2n}x^{n}dx$

9. amistre64

$ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(-1)^{n+1}}{n+1}x^{n+1}$

10. amistre64

forgot the 2^n

11. amistre64
12. amistre64

2^(n+1)

13. amistre64

$ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(-1)^{n+1}2^{n+1}}{n+1}x^{n+1}$ $lim\frac{(-1)^{n+1}2^{n+1}x^{n+1}}{n+1}\frac{n}{(-1)^{n}2^{n}x^{n}}$ $lim\frac{-2x}{n+1}\frac{n}{1}$ $|x|lim\frac{-2n}{n+1} = -2$

14. amistre64

prolly shoulda taken the - with the x :)

15. amistre64

all non - ns should be vacated $|-2x|\ lim\frac{n}{n+1}=2x$ $2x<1;\ x<\frac{1}{2}$