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Chapter 7: Problem 47

Find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{5^{n} n} $$

Short Answer

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

Step by step solution

01

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.
02

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 \(x=2/5\).
03

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 \(\ln(1+\frac{2}{5}) = \ln(\frac{7}{5})\).

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Most popular questions from this chapter

Prove that \(\frac{1}{r}+\frac{1}{r^{2}}+\frac{1}{r^{3}}+\cdots=\frac{1}{r-1}\) for \(|r|>1\).

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty} 6\left(\frac{4}{5}\right)^{n} $$

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{3^{n}}{n^{3}} $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(n>1\), then \(n !=n(n-1) !\)

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