How do you find the power series of ln(1+x)?

Answer 1

Paisley Johnson

#ln(1+x) = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)#

with radius of convergence #R=1#

Start from:

#ln(1+x) = int_0^x (dt)/(1+t)#

Now the integrand function is the sum of a geometric series of ratio #-t#:

#1/(1+t) = sum_(n=0)^oo (-1)^nt^n#

so:

#ln(1+x) = int_0^x sum_(n=0)^oo (-1)^nt^n#

This series has radius of convergence #R=1#, so in the interval #x in (-1,1)# we can integrate term by term:

#ln(1+x) = sum_(n=0)^oo int_0^x (-1)^nt^ndt = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)#

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Answer 2

William Abdalla

To find the power series of ln(1+x), we can start with the known Maclaurin series expansion of ln(1+x), which is:

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...

This series converges for |x| < 1. Therefore, the power series expansion for ln(1+x) is simply the Maclaurin series expansion given above.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.