Proving Convergence of a Series with Positive Terms

[r4nd0m]

hi,
I found this problem in Rudin, and I just can't figure it out.

It goes like this:
Prove that the convergence of \sum a_n a_n \geq 0 implies the convergence of \sum \frac{\sqrt{a_n}}{n}

I tried the comparison test, but that doesn't help because I don't know what the limit \lim_{n \rightarrow \infty} \frac{1}{n\sqrt{a_n}} is equal to.

Then I tried the partial summation formula, \frac{1}{n} \rightarrow \infty and is monotonic, but \sqrt{a_n} > a_n for all but finite many n. \sqrt{a_n} is rising, so if it the partial sums were bounded from above the series would converge, but that isn't true for a_n = \frac{1}{n^2}, 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.

 

[shmoe]

When you tried partial summation, what kind of bounds did you use for \sum_{n\leq x}\sqrt{a_n}?

 

[r4nd0m]

The problem is there are no bounds for it - like in the example I mentioned - \sum a_n = \sum \frac{1}{n^2} converges, but \sum \sqrt{a_n} = \sum \frac{1}{n} doesn't, hence it is also not bounded from above.

 

[shmoe]

I don't mean bounded by a constant for all partial sums, rather an upper bound that depends on x.

With a_n=1/n^2, you can get the bound \sum_{n\leq x}\sqrt{a_n}\leq \log(x)+1. Using this bound, you could go throught the partial summation and show \sum_{n\leq x}\sqrt{a_n}n^{-1} is bounded by a constant.

So, what kind of bounds can you get for \sum_{n\leq x}\sqrt{a_n}? It would be good to ask what kind of bound you'd need to deduce convergence from partial summation as well.

 

[HallsofIvy]

How about the "root test"? Since \Sigma a_n converges, you know that ^n\sqrt{a_n} converges to a number less than or equal to one. What does that tell you about the limit of
{^n\sqrt{\frac\sqrt{a_n}}{n}}= \frac{\sqrt{^n\sqrt{a_n}}}{^n\sqrt{n}}?

 

[r4nd0m]

the root test! I don't know why, but I just ignored it. Thanks a lot

 

[shmoe]

The root test will be inconclusive here if you started with a sequence that has \lim_{n\rightarrow\infty}\sqrt[n]{a_n}=1 to begin with. Or if this limit simply didn't exist (though you can use the lim sup version, again inconclusive if you get a "1").

 

[vivisector]

If <br /> a_n \le {1 \over {n^2 }}<br />, then <br /> \sqrt {a_n } \le {1 \over n} \Rightarrow {{\sqrt {a_n } } \over n} \le {1 \over {n^2 }}<br />;

if <br /> a_n \ge {1 \over {n^2 }}<br /> <br />, then <br /> \sqrt {a_n } \ge {1 \over n}<br /> <br />; multiplying both sides of this inequality by <br /> \sqrt {a_n } <br /> <br /> we obtain <br /> {{\sqrt {a_n } } \over n} \le a_n <br /> <br />.

Thus <br /> {{\sqrt {a_n } } \over n} \le \max \left\{ {a_n ,{1 \over {n^2 }}} \right\} = {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) + {1 \over 2}\left| {a_n - {1 \over {n^2 }}} \right| \le {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) + {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) = a_n + {1 \over {n^2 }}<br /> <br />.

Then the convergence of <br /> \sum\limits_n {a_n } <br /> <br /> and <br /> \sum\limits_n {{1 \over {n^2 }}} <br /> <br /> implies the convergence of

<br /> \sum\limits_n {{{\sqrt {a_n } } \over n}} <br /> <br />.