Exercise · ZW-COMPARISON-TEST-005 · ★☆☆ Basic
Since \(-\frac1n\le\frac1{n^2}\) and \(\sum\frac1{n^2}\) converges, can we conclude that \(\sum\left(-\frac1n\right)\) converges?
Choose one option.
A No: the terms \(-\tfrac1n\) are not nonnegative, so the test does not apply B Yes: it converges, because \(-\tfrac1n\le\tfrac1{n^2}\) and \(\sum\tfrac1{n^2}\) converges C Yes, and its sum is negative and finite D No: because \(\sum\tfrac1{n^2}\) diverges Show answer A. No: the terms \(-\tfrac1n\) are not nonnegative, so the test does not apply← Back to the question
Explanation The test requires \(0\le a_n\). Without it the argument fails: in fact \(\sum\left(-\tfrac1n\right)\) diverges to \(-\infty\).
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