Another series question

1. The problem statement, all variables and given/known data
Prove that if [tex]0 < \alpha < 1 [/tex] and
[tex]\sqrt[n]{|a_n|} <= 1 - \frac{1}{n^{\alpha}}[/tex] for all n >= 1 then the series
[tex]\sum a_n[/tex] converges.


2. Relevant equations



3. The attempt at a solution

First I did:
[tex]|a_n| <= \left({1 - \frac{1}{n^{\alpha}}}\right)^n =
\left({{1 - \frac{1}{n^{\alpha}}}^{n^{\alpha}}}\right)^{n^{1-\alpha}}[/tex]
and since the limit at infinity of [tex]\left({{1-\frac{1}{n^{\alpha}}}^{n^{\alpha}}}\right)[/tex] is
[tex]\frac{1}{e}[/tex] then there exists an N so that for all n>N
[tex]|a_n| <= \left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex]
But how can I show that [tex] \sum \left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex]
converges?
Is there a better way to do this?
Thanks.

 

For the Root Convergence Test you need to have a q<1 so that for all n
[tex] \sqrt[n]{|a_n|} <=q[/tex] but since
[tex]
1 - \frac{1}{n^{\alpha}}
[/tex]
goes to 1 it doesn't apply in this case.

 

Is there another way to approach the problem?
Thanks.

 

Well, [tex] \frac{2}{e} < 1[/tex] and [tex] \frac{1}{n^{\alpha}} > 0[/tex] for all n
and so
[tex] \left({\frac{2}{e}}\right)^{\frac{1}{n^{\alpha}}} < 1[/tex] right?
But the problem is that now r depends on n and r goes to 1 as n goes to infinity so how can you use the geometric series?

 

I tried that and didn't really get any where. I think that maybe
[tex] \sum \left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex]
doesn't even converge. What I have to prove is that
[tex] \left({1 - \frac{1}{n^{\alpha}}}\right)^n [/tex]
converges. Do you have any ideas on how to do that?
Thanks.

 

I started with
[tex] b_n =\left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex] and
[tex] c_n = \left({\frac{2}{e}}\right)^{n}[/tex]
I know that c_n converges so if b_n/c_n goes to some finite number that proves that b_n converges to. But:
[tex] \frac{b_n}{c_n} = \left({\frac{2}{e}}\right)^{n(n^{-\alpha}-1)}[/tex]
and since [tex] \left({\frac{2}{e}}\right)^{n}[/tex] goes to 0 and
[tex] (n^{-\alpha}-1)[/tex] goes to -1 and so b_n/c_n goes to infinity.
Is there any series that's better to compare it to?
And by the way, how did you approximate the sum? Did you simply add up terms up to a "very high"(~1000000) number?
Thanks for you help

 

Then how can I solve the problem if the root test and the comparison test don't work?
Thanks.

 

So what else is there to do to solve the original problem?
Thanks.

 

Do you mean this:
[tex] \frac{1}{1 + e^{{n^p}} + \frac{e^{{n^p}^2}}{2!} + ... +\frac{e^{{n^p}^k} }{k!}...}
[/tex]
?
I don't see how this is easier to evaluate - especially with all the factorials.