find the radius of convergence of the taylor series around x=0 for ln(1/(1+2x))

Question

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

maybe?

amistre64 · Answer

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

amistre64 · Answer

-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)

anonymous · Answer

2 i think

experimentx · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

forgot the 2^n

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5E%28n%2B1%29x%5E%28n%2B1%29%292%5En%2F%28n%2B1%29+from+0+to+inf soo close

amistre64 · Answer

2^(n+1)

amistre64 · Answer

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

amistre64 · Answer

prolly shoulda taken the - with the x :)

amistre64 · Answer

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