Exercise · ZW-COMPARISON-TEST-002 · ★☆☆ Basic
To prove that \(\sum\frac1{n+1}\) diverges, which \(b_n\) of a divergent series gives \(0\le b_n\le\frac1{n+1}\) for \(n\ge1\)?
Choose one option.
Show answer
- A. \(\dfrac{1}{2n}\)
Explanation
Since \(n+1\le2n\), we get \(\frac1{2n}\le\frac1{n+1}\), and \(\sum\frac1{2n}\) diverges. The others converge, so they cannot prove divergence; moreover \(\frac1{n^2}\) is not smaller at \(n=1\).
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
- ZW-COMPARISON-TEST-001 Knowing that ∑1n^2 converges, which series converges by direct comparison with it?
- ZW-COMPARISON-TEST-003 It is known that ∑_n=1^∞12^n=1. Since 12^n+n≤12^n, what upper bound for ∑_n=1^∞12^n+n does the comp…
- ZW-COMPARISON-TEST-004 If 0≤ b_n≤ a_n for all n and ∑ b_n diverges, what happens to ∑ a_n?
- ZW-COMPARISON-TEST-005 Since -1n≤1n^2 and ∑1n^2 converges, can we conclude that ∑(-1n) converges?
- ZW-COMPARISON-TEST-006 Let a_n=(2n+1)/(n^3+4) and b_n=1n^2. Compute lim _n→∞(a_n)/(b_n). (If the limit is finite and posit…
- ZW-COMPARISON-TEST-007 For a_n=(2+cos n)/(3^n) we have 0≤ a_n≤ b_n with b_n=(3)/(3^n). What is ∑_n=1^∞b_n, an upper bound…
- ZW-COMPARISON-TEST-008 Which comparison correctly proves that ∑_n=1^∞13^n-1 converges?
- ZW-COMPARISON-TEST-009 Let a_n≥0 with ∑ a_n convergent. What can be said about ∑ a_n^2?