Question

Find the radius of convergence of the Taylor series around x=0 for
\[\ln (1/(1-7x))\]

Answer 1

can you show some steps,please?

Answer 2

\[\frac 1 {1- 7x}= \sum_{n=1}^\infty (7x)^n
\]
\[\ln(1- 7x)=\int_0^x \frac {dt }{1- 7t}= \sum_{n=1}^\infty \int_0^x (7t)^n dt=
\sum_{n=1}^\infty \frac { (7x)^{n +1}} {n+1}dt
\]

Answer 3

\[ \ln \frac 1 { 1- 7x}= \ln 1 - \ln (1- 7x)= -\ln (1-7x)
\]
From th post above
let \[ c_n = \frac {7^{n+1}} {n+1}
\]
\[\lim_{n\to \infty} \frac {c_n}{c_{n+1}}= \frac 1 7
\]

Answer 4

how do we know the first step?

Answer 5

i guess i understand now

Answer 6

can you kind of explain?

Answer 7

which part?

Answer 8

the first step

Answer 9

1/(1-7x) = summation (7x)^n

Answer 10

this is done using binomial expansion

Answer 11

oh,we didn't cover that!!

Answer 12

(1-7x)^-1 = 1+(-1)/1!*(-7x)+(-1)(-2)/2!(-7x)^2+(-1)(-2)(-3)/3!(-7x)^3+..
i guess it can be done using Tylor series too

Answer 13

yea,i think i got it now!!thanks!!