Taylor Series for ln(1-x): Get Help Now

[teng125]

may i know from ln(1-x) how to become - [infinity (sum) k=1] x^k / k ?
pls help

 

[HallsofIvy]

You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.

 

[VietDao29]

You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.

Uhmm, I think it can also be done with geometric series:
\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1.
So if your final sum is:
\frac{1}{1 - k}, what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).

 

[HallsofIvy]

Uhmm, I think it can also be done with geometric series:
\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1.
So if your final sum is:
\frac{1}{1 - k}, what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).

I thought about that but then integrating is not "pre-calculus" either!