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

Question

Charles Anctil · Accepted Answer

#x^2- x^3/2+x^4/3-...+(-1)^(n-1)x^n/n+..., -1 < x<=1# Power series for #x ln(x+1)# # =x#(power series for # ln(x+1)# #=x(x-x^2/2+x^3/3-...), -1< x<=1# #x^2-x^3/2+x^4/3-...+(-1)^(n-1)x^n/n+..., -1 < x<=1#

William Abdalla · Answer

undefined To find a power series representation for $f(x) = x \ln(x + 1)$, we can start by recognizing that $\ln(x + 1)$ has a known power series representation:

$$\ln(1 + z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots$$

Substitute $z = x$ into this series, then multiply the resulting series by $x$. This gives us the power series representation for $x \ln(x + 1)$.

The radius of convergence of this series can be found using the ratio test:

$$R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right|$$

Where $a_{n}$ is the coefficient of the nth term in the power series representation.