Let a_n≥0 with ∑ a_n convergent. What can be said about ∑ a_n^2?

It converges: since a_n→0, for large n we have a_n^2≤ a_n and we compare with ∑ a_n

It converges only if (a_n) is decreasing

Exercise · ZW-COMPARISON-TEST-009 · ★★★ Advanced

Let \(a_n\ge0\) with \(\sum a_n\) convergent. What can be said about \(\sum a_n^2\)?

Choose one option.

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A. It converges: since \(a_n\to0\), for large \(n\) we have \(a_n^2\le a_n\) and we compare with \(\sum a_n\)

Explanation

If \(\sum a_n\) converges, \(a_n\to0\), so there is \(N\) with \(a_n\le1\) for \(n\ge N\) and then \(0\le a_n^2\le a_n\). Comparison gives convergence, with no monotonicity needed.

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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 \(a_n\le b_n\).

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

Let \(a_n\ge0\) with \(\sum a_n\) convergent. What can be said about \(\sum a_n^2\)?

Explanation

Review the concept: Comparison test

More practice problems: comparison test