Problem 48 - 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. $$ \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2 n+1} $$

Step-by-Step Solution

Answer

The sum of the series \(\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2 n+1}\) is \(\frac{\pi}{4}\) or approximately 0.785.

Step 1: Identify the series and function type

The given series \(\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2 n+1}\) is a form of an alternating series where the terms alternate in sign from positive to negative. The function form is \((-1)^{n} \frac{1}{2 n+1}\) where the denominator changes with each term \(n\).

Step 2: Identify a similar well-known function

The well-known function this series is similar to is \(\arctan(x) = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+1}\). We can see the similarity as the general formulas for both series match.

Step 3: Calculate the sum of the series

Since \((-1)^{n} \frac{x^{2n+1}}{2n+1}\) evaluated at \(x = 1\) is \((-1)^{n} \frac{1}{2 n+1}\), the sum of the series will be \(\arctan(1)\). The inverse tangent of 1 is \(\frac{\pi}{4}\), which is approximately 0.785.