Problem 46 - Essential Calculus

Question

Find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{n} n} $$

Step-by-Step Solution

Answer

The sum of the series is \(ln(4/3)\).

Step 1: Identify the Taylor series

The Taylor series expansion of the natural logarithm function ln(1+x) is given by \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^n}{n}\]. Compare this with the given series \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^n n}\]. Notice that if we let x=1/3 in the Taylor series expansion, we'd get the original series.

Step 2: Apply suitable function

Substitute x=1/3 into the natural logarithmic function ln(1+x), to get \[ln(1+1/3)=ln(4/3)\]. This is the sum of the original series.