Exercise · ZW-COMPARISON-TEST-007 · ★★☆ Intermediate

For \(a_n=\dfrac{2+\cos n}{3^n}\) we have \(0\le a_n\le b_n\) with \(b_n=\dfrac{3}{3^n}\). What is \(\displaystyle\sum_{n=1}^{\infty}b_n\), an upper bound for \(\sum a_n\)?

Choose one option.

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  • B. \(\tfrac32\)

Explanation

Since \(2+\cos n\le3\), take \(b_n=3\cdot3^{-n}\): its sum is \(3\cdot\frac{1/3}{1-1/3}=\tfrac32\).

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Review the concept: Comparison test

A test that decides whether a series \(\sum a_n\) of nonnegative terms converges by comparing it with another series \(\sum b_n\) such that \(a_n\le b_n\).

\[0\le a_n\le b_n\ \Longrightarrow\ \Big(\sum b_n\text{ converges}\Rightarrow\sum a_n\text{ converges}\Big)\]

More practice problems: comparison test

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