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91Ó°ÊÓ

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n}}\) always convergent? Either prove it or give a counterexample.

Short Answer

Expert verified
The statement is false. A counterexample is the series \(a_{n} = \frac{1}{n^2}\) (which is convergent) and its square root series \(\sum \sqrt{a_{n}} = \sum {\frac{1}{n}}\) which is divergent.

Step by step solution

01

Understanding the problem statement

We have a series \(\sum a_{n}\) which we know is convergent, i.e., its partial sums form a sequence that has a finite limit. The problem asks if the series \(\sum \sqrt{a_{n}}\) is always convergent given this.
02

Formulating a counterexample

Consider a series whose terms are \(a_{n} = \frac{1}{n^2}\). We know this series to be convergent as it is a p-series with p=2 which is greater than 1. Now, consider \(\sum \sqrt{a_{n}} = \sum {\frac{1}{n}}\). This is a p-series with p=1 and is known to be divergent (harmonic series). This serves as a counterexample, disproving the statement that \(\sum \sqrt{a_{n}}\) is always convergent if \(\sum a_{n}\) is.
03

Final Conclusion

The series \(\sum \sqrt{a_{n}}\) is not always convergent even if the original series \(\sum a_{n}\) is convergent. A counterexample exists to disprove the claim (for instance, the series of \(\frac{1}{n^2}\) and its square root series). Therefore, convergence of an original series does not guarantee the convergence of its corresponding square root series.

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