The Maclaurin series for $\ln\left(\dfrac{1}{1-x}\right)$ is $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ with interval of convergence $-1 \leq x < 1$. (a) Find the Maclaurin series for $\ln\left(\dfrac{1}{1+3x}\right)$ and determine the interval of convergence. (b) Find the value of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$. (c) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ converges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ diverges. Give reasons why your value of $p$ is correct. (d) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ converges. Give reasons why your value of $p$ is correct.
The Maclaurin series for $\ln\left(\dfrac{1}{1-x}\right)$ is $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ with interval of convergence $-1 \leq x < 1$.
(a) Find the Maclaurin series for $\ln\left(\dfrac{1}{1+3x}\right)$ and determine the interval of convergence.
(b) Find the value of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$.
(c) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ converges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ diverges. Give reasons why your value of $p$ is correct.
(d) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ converges. Give reasons why your value of $p$ is correct.