11. -/3 points details my notes scalcet9 11.9.029.ep. find a power series representation for the function…

Explanation:

Step1: Recall power - series of $\ln(1 + x)$

The power - series of $\ln(1 + x)=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n}$, for $|x|\lt1$.

Step2: Find power - series of $g(x)=x^{5}\ln(1 + x)$

Multiply the power - series of $\ln(1 + x)$ by $x^{5}$: $g(x)=x^{5}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n}=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n + 5}$.

Step3: Evaluate the indefinite integral of $x^{5}\ln(1 + x)$

$\int x^{5}\ln(1 + x)dx=\int\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n + 5}dx$. Integrate term - by - term: $\int\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n + 5}dx=C+\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n(n + 6)}x^{n+6}$.

Step4: Find the radius of convergence

We know that the power - series of $\ln(1 + x)$ has a radius of convergence $R = 1$. When we multiply by $x^{5}$ and integrate term - by - term, the radius of convergence remains the same. So the radius of convergence $R = 1$.

Answer:

$g(x)=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}x^{n + 5}$; $\int x^{5}\ln(1 + x)dx=C+\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n(n + 6)}x^{n+6}$; $R = 1$