Convergence of Series with Square Root Terms

[sutupidmath]

Hi all,

A friend of mine asked me if i had any ideas about the following problem, i tackled it but with no success, so i thought i would post it here.
It is not a homework problem, or a regular textbook problem.

Problem:

If we know that a series with positive terms :

\sum_{i=1}^{\infty}a_i<\infty

Then what can we say about the nature of the series:

\sum_{i=1}^{\infty}\sqrt{\frac{a_i}{i}}

?


Thanks!

 

[g_edgar]

Try two examples:
a_i = \frac{1}{i^2}
and
a_i = \frac{1}{i(\log (i+1))^2}

 

[sutupidmath]

I appreciate your input. I should've thought about the second example><

 

[╔(σ_σ)╝]

I'm not sure of how to show this but my mind tells me the second series converges.

My reasoning is as follows:

Since the first series converges you can find a p series that is relatively close to a_i ( for example i^(1.06) from the p-series we know that 1/i^(p) converges for p >1).

Assuming your original series if approximated by a p-series that conveges say 1/ (i^p) we can say that the second series is approximately sigma ( 1/ n^((p+1)/2)) ,which converges since P>1, and p+1 > 2 the second series becomes a p-series with p >1 also.

Thus convergence can be established.

Unfortunately I can not think of any convergence test that would establish convergence more clearly and nicely. There are probably some test out there that you can use to establish convegence. You may even be to establish some nice inequality that would make convergence more obvious and mathematically sound. I was thinking of the Cauchy -Schwartz but i don't see any immediate benefit of using it.

 

[sutupidmath]

This is a counter example:

a_i = \frac{1}{i(\log (i+1))^2}

Since the first series converges while the second diverges.

Case closed!

 

[Office_Shredder]

I appreciate your input. I should've thought about the second example><

It's a pretty sophisticated counterexample, don't think it should have been obvious.

 

[g_edgar]

It's a pretty sophisticated counterexample, don't think it should have been obvious.

It is a standard example for "the integral test" where ratio and root tests fail. If you don't know it, then probably your textbook will do "the integral test" later.

 

[╔(σ_σ)╝]

This is a counter example:

a_i = \frac{1}{i(\log (i+1))^2}

Since the first series converges while the second diverges.

Case closed!

Sorry I didn't look at that post.

My mistake lol.

I guess I was wrong about convergence.lol