Problem 49 - 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. $$ \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2^{2 n+1}(2 n+1)} $$

Step-by-Step Solution

Answer

The sum of the series is approximately 0.245, which is the result of \( Arctan(\frac{1}{4}) \).

Step 1: Identify The Series Structure

This series resembles the Taylor Series expansion of the Arctan function: \( Arctan(x) = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n+1}}{2n+1} \). In our given series, x is replaced by \( \frac{1}{2^2} \) or \( \frac{1}{4} \). So, we can rewrite the series as \( \sum_{n=0}^{\infty}(-1)^{n} \frac{(1/4)^{n+1}}{2n+1} \).

Step 2: Apply Known Function To Series

By comparing the series with the Taylor series of the Arctan function, it is known that our series is a specific representation of the Arctan function that applies when \( x = \frac{1}{4} \). Therefore, calculating the sum of the series equates to evaluating \( Arctan(\frac{1}{4}) \).

Step 3: Calculate The Function

The result can be found in any scientific calculator or a tool such as Wolfram Alpha, this gives the result of \( Arctan(\frac{1}{4}) \approx 0.245 \).