Problem 47 - 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. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{5^{n} n} $$

Step-by-Step Solution

Answer

The sum of the series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{5^{n} n} \) is \(\ln(\frac{7}{5})\).

Step 1: Recognize the Series

Look at the series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{5^{n} n}\). The factor \((-1)^{n+1}\) implies that the series is alternating. Also, the terms \( \frac{2^{n}}{5^{n} n} \) suggest that it is a variation of geometric series.

Step 2: Identify the Well-known Function

The series looks like the Taylor series for the natural logarithm, which is \(\ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{n}}{n}\). The \( \frac{x^{n}}{n} \) part matches with our series when \(x=2/5\), so the well-known function that we can use to represent the series is \(\ln(1+x)\) with \(x=2/5\).

Step 3: Compute the Sum

Substitute \(x=2/5\) into the formula for the sum of the series \(\ln(1+x)\). This will give the sum of the series to be \(\ln(1+\frac{2}{5}) = \ln(\frac{7}{5})\).