# Series problem

1. Oct 3, 2006

### 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.

Last edited: Oct 3, 2006
2. Oct 3, 2006

### shmoe

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

3. Oct 4, 2006

### 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.

Last edited: Oct 4, 2006
4. Oct 4, 2006

### 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.

Last edited: Oct 4, 2006
5. Oct 4, 2006

### HallsofIvy

Staff Emeritus
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
$${^n\sqrt{\frac\sqrt{a_n}}{n}}= \frac{\sqrt{^n\sqrt{a_n}}}{^n\sqrt{n}}$$?

6. Oct 4, 2006

### r4nd0m

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

7. Oct 4, 2006

### 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").

8. May 4, 2008

### vivisector

If $$a_n \le {1 \over {n^2 }}$$, then $$\sqrt {a_n } \le {1 \over n} \Rightarrow {{\sqrt {a_n } } \over n} \le {1 \over {n^2 }}$$;

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

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

Then the convergence of $$\sum\limits_n {a_n }$$ and $$\sum\limits_n {{1 \over {n^2 }}}$$ implies the convergence of

$$\sum\limits_n {{{\sqrt {a_n } } \over n}}$$.