Another series question

[daniel_i_l]

Homework Statement

Prove that if $$0 < \alpha < 1$$ and
$$\sqrt[n]{|a_n|} <= 1 - \frac{1}{n^{\alpha}}$$ for all n >= 1 then the series
$$\sum a_n$$ converges.

Homework Equations

The Attempt at a Solution

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

 

[e(ho0n3]

How about using the Root Convergence Test?

 

[daniel_i_l]

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

 

[daniel_i_l]

Is there another way to approach the problem?
Thanks.

 

[e(ho0n3]

But how can I show that $$\sum \left({\frac{2}{e}}\right)^{n^{1-\alpha}}$$
converges?
Is there a better way to do this?

How about this: Write

$$\left({\frac{2}{e}}\right)^{n^{1-\alpha}} = r^n$$

where

$$r = \left({\frac{2}{e}}\right)^{1/n^\alpha}$$

If you can show that |r| < 1, then you can use the geometric series.

 

[daniel_i_l]

Well, $$\frac{2}{e} < 1$$ and $$\frac{1}{n^{\alpha}} > 0$$ for all n
and so
$$\left({\frac{2}{e}}\right)^{\frac{1}{n^{\alpha}}} < 1$$ 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?

 

[e(ho0n3]

Oops. That's right. So how about using a comparison test with a geometric series?

 

[daniel_i_l]

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

 

[e(ho0n3]

I tried that and didn't really get any where. I think that maybe
$$\sum \left({\frac{2}{e}}\right)^{n^{1-\alpha}}$$
doesn't even converge.

I made a perl script to compute the sum with alpha = 0.5. The sum converges to about 20.8.

Show us what you did with the comparison test.

 

[daniel_i_l]

I started with
$$b_n =\left({\frac{2}{e}}\right)^{n^{1-\alpha}}$$ and
$$c_n = \left({\frac{2}{e}}\right)^{n}$$
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:
$$\frac{b_n}{c_n} = \left({\frac{2}{e}}\right)^{n(n^{-\alpha}-1)}$$
and since $$\left({\frac{2}{e}}\right)^{n}$$ goes to 0 and
$$(n^{-\alpha}-1)$$ 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

 

[e(ho0n3]

And by the way, how did you approximate the sum? Did you simply add up terms up to a "very high"(~1000000) number?

Yes I did.

 

[daniel_i_l]

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

 

[e(ho0n3]

I've realized that comparing to a geometric series won't work because there will always be a term greater than the corresponding term in the geometric series.

 

[daniel_i_l]

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

 

[e(ho0n3]

I've been looking at other convergence tests but I don't see one that would work. This one is tough.

 

[Dick]

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

Ok, try this. Show (1-x/n)^n<=exp(-x) for 0<x<n. I think it's true. Prove it. Then if all goes well, you can compare you original series with exp(-n^p) where p=1-alpha. So 0<p<1. That also fails the root test and the ratio test, etc. That probably means we are on the right track. Series expand the exponential in 1/exp(n^p). Notice the increasing powers of n^p? At some point you reach a point where (n^p)^k=n^(p*k) is such that p*k>1. Now what??

 

[e(ho0n3]

$(1 - \frac{x}{n})^n \le e^{-x}$ is equivalent to

(*) n ln(1 - x/n) ≤ -x

As $x \to 0$, (*) becomes 0 ≤ 0. As $x \to n$, (*) becomes $-\infty \le -n$. Both sides are decreasing in the interval (0, n) but the LHS seems to be decreasing faster. This is conclusively shown by comparing the derivatives of n ln(1 - x/n) with -x.

At some point you reach a point where (n^p)^k=n^(p*k) is such that p*k>1.

Why are you concluding that? It seems to be true when k = 1, but then pk < 1. What is the point of this?

 

[Dick]

I know you are interested, but try not to take over from daniel_i_l and work on his problem for him, ok? I.e. stop showing the mechanics of something I left as a routine exercise. But if your working is correct then that would work for a proof of the first one. I didn't do it that way and I'm not going to check it. For the second part the strategy is to compare with a p-series. So I want an exponent bigger than one. Enuf said?

 

[e(ho0n3]

Oops. Sorry about that. I don't want to take over someone else's thread.

 

[daniel_i_l]

Series expand the exponential in 1/exp(n^p). Notice the increasing powers of n^p? At some point you reach a point where (n^p)^k=n^(p*k) is such that p*k>1. Now what??

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

 

[Dick]

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

Great. You waited for a month and I lost my train of thought. You aren't trying to evaluate it, you are just trying to show it converges. The point (I think) was 1/e^n definitely converges. If the exponent of e is larger than n then it also converges. The factorials are just constants. Think comparison test.

 

[daniel_i_l]

I thought of another way to do it:
since you know that $$\left({1 - \frac{1}{n^{\alpha}}}\right)^{n^{\alpha}}$$ goes to 1/e then there exists a c so that 1/e < c/e < 1/2 (since 2<e) and an N so that for all n>N
$$\left({1 - \frac{1}{n^{\alpha}}}\right)^{n^{\alpha}} < \frac{c}{e}$$ and then for all n>N
$$|a_n| <= \left({\frac{c}{e}}\right)^{n^{\beta}}$$
Noe we know that if the sequence a_n is decreasing and bigger than 0 then the corresponding series converges iff the series corresponding to $$2^n a_{2^n}$$ converges. So in our case we check $$2^n \left({\frac{c}{e}}\right)^{2^{n \beta}}$$ Now there exists an N do that for all n>N $$2^{n \beta} > n$$ and then
$$2^n \left({\frac{c}{e}}\right)^{2^{n \beta}} <= 2^n \left({\frac{c}{e}}\right)^n} = \left({\frac{2c}{e}}\right)^n}$$ But 2c/e < 1 and so the series converges and so from the comparison test the series of a_n converges.
What do you think?

 

[Dick]

I didn't check all of the details, but yes. It seems something like that should also work.