- #1
r4nd0m
- 96
- 1
hi,
I found this problem in Rudin, and I just can't figure it out.
It goes like this:
Prove that the convergence of [tex]\sum a_n[/tex] [tex] a_n \geq 0[/tex] implies the convergence of [tex] \sum \frac{\sqrt{a_n}}{n} [/tex]
I tried the comparison test, but that doesn't help because I don't know what the limit [tex]\lim_{n \rightarrow \infty} \frac{1}{n\sqrt{a_n}}[/tex] is equal to.
Then I tried the partial summation formula, [tex] \frac{1}{n} \rightarrow \infty [/tex] and is monotonic, but [tex]\sqrt{a_n} > a_n[/tex] for all but finite many n. [tex]\sqrt{a_n}[/tex] is rising, so if it the partial sums were bounded from above the series would converge, but that isn't true for [tex] a_n = \frac{1}{n^2} [/tex], so I can't use this way.
The last thing that comes to my mind is to use the Cauchy criterion, but I can't find any good use of it here. a_n will be smaller than any epsilon for infinitely many n, but that doesn't really help.
Have I missed something out, or done something in a wrong way? Thanks for any help.
I found this problem in Rudin, and I just can't figure it out.
It goes like this:
Prove that the convergence of [tex]\sum a_n[/tex] [tex] a_n \geq 0[/tex] implies the convergence of [tex] \sum \frac{\sqrt{a_n}}{n} [/tex]
I tried the comparison test, but that doesn't help because I don't know what the limit [tex]\lim_{n \rightarrow \infty} \frac{1}{n\sqrt{a_n}}[/tex] is equal to.
Then I tried the partial summation formula, [tex] \frac{1}{n} \rightarrow \infty [/tex] and is monotonic, but [tex]\sqrt{a_n} > a_n[/tex] for all but finite many n. [tex]\sqrt{a_n}[/tex] is rising, so if it the partial sums were bounded from above the series would converge, but that isn't true for [tex] a_n = \frac{1}{n^2} [/tex], so I can't use this way.
The last thing that comes to my mind is to use the Cauchy criterion, but I can't find any good use of it here. a_n will be smaller than any epsilon for infinitely many n, but that doesn't really help.
Have I missed something out, or done something in a wrong way? Thanks for any help.
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