attempt 1: 10 attempts remaining. given that ∑(n = 0)^∞ x^n = 1 / (1 - x), for |x| < 1, use term by term…

attempt 1: 10 attempts remaining. given that ∑(n = 0)^∞ x^n = 1 / (1 - x), for |x| < 1, use term by term integration to find the power series for f(x) = ln(4 - x) centered at a = 0. you may need an intial step to manipulate the given series before integrating. the power series for f(x) centered at 0 is ln(4) - ∑(n = 0)^∞ help (formulas)

attempt 1: 10 attempts remaining. given that ∑(n = 0)^∞ x^n = 1 / (1 - x), for |x| < 1, use term by term integration to find the power series for f(x) = ln(4 - x) centered at a = 0. you may need an intial step to manipulate the given series before integrating. the power series for f(x) centered at 0 is ln(4) - ∑(n = 0)^∞ help (formulas)

Answer

Explanation:

Step1: Rewrite the function

First, rewrite $\ln(4 - x)$ as $\ln\left[4\left(1-\frac{x}{4}\right)\right]=\ln(4)+\ln\left(1 - \frac{x}{4}\right)$.

Step2: Recall the given series

We know that $\sum_{n = 0}^{\infty}x^{n}=\frac{1}{1 - x}$ for $|x|<1$. Replace $x$ with $\frac{x}{4}$, we get $\sum_{n = 0}^{\infty}\left(\frac{x}{4}\right)^{n}=\frac{1}{1-\frac{x}{4}}$ for $\left|\frac{x}{4}\right|<1$ (i.e., $|x|<4$).

Step3: Integrate term - by - term

We know that $\ln(1 - t)=-\int\frac{1}{1 - t}dt$. Let $t=\frac{x}{4}$, then $\frac{1}{1-\frac{x}{4}}=\sum_{n = 0}^{\infty}\left(\frac{x}{4}\right)^{n}$. Integrating both sides with respect to $x$: [ \begin{align*} \int\frac{1}{1-\frac{x}{4}}dx&=\int\sum_{n = 0}^{\infty}\left(\frac{x}{4}\right)^{n}dx\ -\ln\left(1-\frac{x}{4}\right)&=\sum_{n = 0}^{\infty}\frac{1}{n + 1}\left(\frac{x}{4}\right)^{n+1}+C \end{align*} ] When $x = 0$, $-\ln(1)=C$, so $C = 0$. Then $\ln\left(1-\frac{x}{4}\right)=-\sum_{n = 0}^{\infty}\frac{x^{n + 1}}{(n+1)4^{n+1}}$.

Answer:

$\sum_{n = 0}^{\infty}\frac{x^{n + 1}}{(n + 1)4^{n+1}}$