Exercise · ZW-COMPARISON-TEST-003 · ★☆☆ Basic

It is known that \(\sum_{n=1}^{\infty}\frac1{2^n}=1\). Since \(\frac1{2^n+n}\le\frac1{2^n}\), what upper bound for \(\sum_{n=1}^{\infty}\frac1{2^n+n}\) does the comparison give directly?

Choose one option.

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  • C. \(1\)

Explanation

The majorant sums to \(1\), so the series converges and its sum is \(\le1\).

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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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