Testing the Convergence of Series: A Counterexample

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SUMMARY

The statement "If \(\sum_{k=1}^{\infty} a_k\) converges and \(\frac{a_k}{b_k} \to 0\) as \(k \to \infty\), then \(\sum_{k=1}^{\infty} b_k\) converges" is false. A valid counterexample is given with \(a_k = 0\) and \(b_k = \frac{1}{k}\), which meets the conditions of convergence for \(a_k\) while \(b_k\) diverges. An alternative, less trivial counterexample is \(a_k = \frac{1}{k^2}\), which also demonstrates the same divergence of \(\sum_{k=1}^{\infty} b_k\).

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Homework Statement



If [tex]\sum_{k=1}^{\infty} a_k[/tex] converges and [tex]a_k/b_k \to 0[/tex] as [tex]k\to \infty[/tex], then [tex]\sum_{k=1}^{\infty} b_k[/tex] converges.

Homework Equations


It is true or false.

The Attempt at a Solution


I think it is false and here is my counterexample. Let [tex]a_k = 0,b_k=\frac{1}{k}[/tex]. This satisfies our initial conditions of [tex]\sum_{k=1}^{\infty} a_k[/tex] converges and [tex]a_k/b_k \to 0[/tex] as [tex]k\to \infty[/tex] but [tex]\sum_{k=1}^{\infty} b_k[/tex] diverges.
Is this correct?
 
Last edited:
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Looks okay.

Your counterexample also looks correct, if you want to make it slightly less trivial you could use
[tex]a_k = \frac{1}{k^2}[/tex]
instead :)
 

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