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Chapter 7: Problem 43

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{3^{n}+1} $$

Short Answer

Expert verified
The series \(\sum_{n=0}^{\infty} \frac{1}{3^{n}+1}\) converges.

Step by step solution

01

Identify a known series for comparison

Since the general term in the series is \(\frac{1}{3^n +1}\), choosing the series \(\sum_{n=0}^{\infty}\frac{1}{3^n}\) for comparison purposes is reasonable. The convergence of this series is known because it is a geometric series with a common ratio \(r = \frac{1}{3}\) where \(|r|<1\). Hence, it converges.
02

Apply the Direct Comparison Test

To apply the Direct Comparison Test, we must show that the terms of our series are always less than or equal to the terms of the convergent series we've chosen for comparison. We have \(\frac{1}{3^n+1}\) and \(\frac{1}{3^n}\). Since we're adding 1 in the denominator of the former, it will always be less than or equal to the latter for all n, so we have \(\frac{1}{3^n+1} \leq \frac{1}{3^n}\). This inequality proves that the terms of our series are less compared to the corresponding terms of the converging series.
03

Conclude with the result

Since the terms of our series are always less than or equal to the terms of a known converging series, by the Direct Comparison Test, the original series \(\sum_{n = 0}^{\infty}\frac{1}{3^n+1}\) must also converge.

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Most popular questions from this chapter

An electronic games manufacturer producing a new product estimates the annual sales to be 8000 units. Each year, \(10 \%\) of the units that have been sold will become inoperative. So, 8000 units will be in use after 1 year, \([8000+0.9(8000)]\) units will be in use after 2 years, and so on. How many units will be in use after \(n\) years?

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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.
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