Question

Find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{2 n-1}(2 n-1)} $$

Short answer

The sum of the series is \( \arctan(\frac{1}{3}) \)

Step-by-step solution

Identify the well-known function

The given series looks similar to the Taylor series expansion of \( \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots \) .

Draw parallels with Taylor series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{2 n-1}(2 n-1)} \), which can be rewritten as \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(2 n-1)3^{2 n-1}} \). This is very similar to \( \arctan(x) \) except that every x in \( \arctan(x) \) is replaced by \( \frac{1}{3} \) in the given series.

Use known function to find sum

Thus, we see that the given series is merely the series expansion of \( \arctan(x) \) evaluated at \( x = \frac{1}{3} \). Therefore, the sum of the given series equals \( \arctan(\frac{1}{3}) \).