Exercise · ZW-COMPARISON-TEST-010 · ★★★ Advanced

Using \(n!\ge2^{n-1}\) for \(n\ge1\), what upper bound for \(\sum_{n=1}^{\infty}\frac1{n!}\) comes from summing the geometric series \(\sum_{n=1}^{\infty}\frac1{2^{n-1}}\)?

Choose one option.

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  • D. \(2\)

Explanation

Then \(\frac1{n!}\le\frac1{2^{n-1}}\) and \(\sum_{n=1}^{\infty}2^{-(n-1)}=\frac{1}{1-1/2}=2\). The series converges and its sum is \(\le2\).

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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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