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Chapter 7: Problem 46

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

Short Answer

Expert verified
The sum of the series is \(ln(4/3)\).

Step by step solution

01

Identify the Taylor series

The Taylor series expansion of the natural logarithm function ln(1+x) is given by \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^n}{n}\]. Compare this with the given series \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^n n}\]. Notice that if we let x=1/3 in the Taylor series expansion, we'd get the original series.
02

Apply suitable function

Substitute x=1/3 into the natural logarithmic function ln(1+x), to get \[ln(1+1/3)=ln(4/3)\]. This is the sum of the original series.

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Most popular questions from this chapter

Show that the series \(\sum_{n=1}^{\infty} a_{n}\) can be written in the telescoping form \(\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_{n}\right)\right]\) where \(S_{0}=0\) and \(S_{n}\) is the \(n\) th partial sum.

State the definitions of convergent and divergent series.

Let \(\left\\{a_{n}\right\\}\) be a monotonic sequence such that \(a_{n} \leq 1\). Discuss the convergence of \(\left\\{a_{n}\right\\} .\) If \(\left\\{a_{n}\right\\}\) converges, what can you conclude about its limit?

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(1+\frac{k}{n}\right)^{n} $$

In your own words, define each of the following. (a) Sequence (b) Convergence of a sequence (c) Monotonic sequence (d) Bounded sequence
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