Question

The Maclaurin series for ln(1+x) is given by

Answer 1

(a)

Replacing x with x/3 and then multiplying both sides by x

$\begin{aligned} \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots + (-1)^{n+1} \frac{x^n}{n} + \cdots \\ \\ \ln\left(1 + \tfrac{x}{3}\right) &= \frac{x}{3} - \frac{x^2}{3^2 \cdot 2} + \frac{x^3}{3^3 \cdot 3} - \frac{x^4}{3^4 \cdot 4} + \cdots + (-1)^{n+1} \frac{x^n}{3^n \cdot n} \\ \\ x\ln\left(1 + \tfrac{x}{3}\right) &= \frac{x^2}{3} - \frac{x^3}{3^2 \cdot 2} + \frac{x^4}{3^3 \cdot 3} - \frac{x^5}{3^4 \cdot 4} + \cdots + (-1)^{n+1} \frac{x^{n+1}}{3^n \cdot n} \end{aligned}$

The general term is

$\displaystyle(-1)^{n+1} \frac{x^{n+1}}{3^n \cdot n}$

____________

(b)

Due to the powers present in the general term, use the ratio test

$\displaystyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left| \frac{ \frac{x^{n+2}}{3^{n+1}(n+1) } }{\frac{x^{n+1}}{3^n n} } \right| = \lim_{n\to\infty}\left|\frac{xn}{3n+3} \right| = \left|\frac{x}{n} \right| \ \textless \ 1$

The series converges on x ∈ (-3,3)
Determining the interval of convergence, test the endpoints

For x = -3, general term becomes

$\displaystyle \frac{(-1)^{n+1}(-3)^{n+1}}{3^n \cdot n} \\ \\ = \frac{(-1)^{n+1}(-1)^{n+1}(3)^{n+1}}{3^n \cdot n} \\ \\ = (-1)^{2n+2} \cdot \frac{3}{n}$

$(-1)^{2n+2} \implies 1 \text{ because 2n+2 is all even}$

$\sum_{n=1}^{\infty} \frac{3}{n} \text{ diverges via harmonic series}$

Diverges at x = -3.

For x = -3, general term becomes

$\displaystyle \frac{(-1)^{n+1}3^{n+1}}{3^n \cdot n} = (-1)^{n+1} \frac{3}{n} \\ \\ \implies 3 \sum_{n=1}^{\infty} \frac{3}{n}$

This is an alternate harmonic series, so it converges as the terms are decreasing in size (approaching zero as n approaches infinity)

The interval of convergence is

-3 < x ≤ 3

____________

(c)

We wrote all the terms above in part (a). The neglected term in the alternating series of f is

$\displaystyle - \frac{x^5}{3^4 \cdot 4}$

Thus

$\displaystyle \left| P_4(2) - f(2) \right| \le \left| - \frac{2^5}{3^4 \cdot 4} \right| = \frac{8}{81}$