How do you find a power series representation for #f(x)=ln(1+x)# and what is the radius of convergence?

Question

Emily Ammerman · Accepted Answer

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

Start from the geometric series:

#sum_(n=0)^oo q^n = 1/(1-q)#

converging for #abs q < 1#.

Let #q=-t#, then:

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

converging for #abs t < 1#

Inside the interval of convergence, that is for #t in (-1,1)#, we can integrate term by term:

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

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

obtaining a series with at least the same radius of convergence #R=1#.

Adam Adkins · Answer

undefined To find the power series representation for $f(x) = \ln(1+x)$, we can use the known power series expansion for $\ln(1+x)$:

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

The radius of convergence for this power series can be found using the ratio test. The ratio test states that if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L $, then the radius of convergence is $ R = \frac{1}{L} $, where $ L $ is the limit.

Applying the ratio test to the power series for $ \ln(1 + x) $:

$$ \lim_{n \to \infty} \left| \frac{(-1)^{n+2} \frac{(x^{n+1})}{n+1}}{(-1)^{n+1} \frac{x^n}{n}} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{-x^{n+1} (n)}{(n+1)(x^n)} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{-x}{n+1} \right| $$

$$ = 0 $$

Thus, the limit of the absolute value of the ratio of consecutive terms is 0, indicating that the radius of convergence $ R $ is infinite. Therefore, the power series representation for $ f(x) = \ln(1+x) $ converges for all real numbers $ x $.