Convergence of Series using the Integral Test

[llooppii]

Homework Statement

Determine whether the series (from 1 to infinity) \Sigma 1/(n^(1+1/n)) converges.

The Attempt at a Solution

I know that since it is positive, continuous, and decreasing, I am able to use the integral test, but I don't know how to anti-differentiate that. I've tried a u-sub, but that didn't lead anywhere. Any help is much appreciated!

 

[micromass]

I would first try Cauchys condensation test. It says that

\sum{u_n}~\text{converges if and only if}~\sum{2^nu_{2^n}}~\text{converges}

Tell me if you don't know this or if it is inconclusive. I'll figure something else out...

 

[llooppii]

I don't know that test. How does it work?

 

[micromass]

Well, you take a series with term u_n. You change this into a series with term 2^nu_{2^n]. If this series converges/diverges, then so does the original series.

Example:
The series \sum{1/n} diverges since the series \sum{2^n/2^n} diverges.

Okay, let's try something else. Do you know the following test?
If {a_n}, {b_n} > 0 and if \lim_{n\rightarrow +\infty}{\frac{a_n}{b_n}} exists and is nonzero, then

\sum_{n=1}^{+\infty}~\text{converges iff}~ \sum_{n=1}^{+\infty}~\text{converges}


If not, can you list all the tests you've seen?

 

[micromass]

Also, try a ratio test. I think that might work.
An integral test, like you tried, is hard cause u don't know that integral...

 

[Dick]

Homework Statement

Determine whether the series (from 1 to infinity) \Sigma 1/(n^(1+1/n)) converges.

The Attempt at a Solution

I know that since it is positive, continuous, and decreasing, I am able to use the integral test, but I don't know how to anti-differentiate that. I've tried a u-sub, but that didn't lead anywhere. Any help is much appreciated!

You can't apply the integral test directly. Your series is 1/(n*n^(1/n)). Try to think of a function you can replace the n^(1/n) part with that allows you to integrate it and then apply a comparison test. Hint: your given series is really close to being 1/n.

 

[llooppii]

Ah thank you! That really helped. Could i just say that because n^(1+1/n) is always greater than n, that the series must converge because p-series states that p > 1, then the series converges??

 

[Dick]

Ah thank you! That really helped. Could i just say that because n^(1+1/n) is always greater than n, that the series must converge because p-series states that p > 1, then the series converges??

No. Because it's not a p-series. I'll also give you the hint that it diverges. Can you find a series that's smaller than the given series that you can show diverges?

 

[micromass]

No, because p=1+1/n and this is dependent of n. We can only say that 1/n^p converges if p is constant. I'd still try the ration test...

 

[llooppii]

i really can't think of anything that works because i always end up having n^n for i don't know a test for...

 

[Dick]

No, because p=1+1/n and this is dependent of n. We can only say that 1/n^p converges if p is constant. I'd still try the ration test...

The ratio test gives you 1. Did you try it?

 

[Dick]

i really can't think of anything that works because i always end up having n^n for i don't know a test for...

What do you think about 1/(n*log(n))? Convergent, divergent?

 

[micromass]

Does following equation help you?

\frac{1}{2n}\leq \frac{1}{n^{1+1/n}}

 

[llooppii]

i must have done the ratio test wrong then, but it would still fail if it equals 1 though, rite? and I am not too sure about nlogn, but i definitely see how 1/2n works. thank you very much micromass and Dick, i really appreciate your time!

 

[micromass]

Yeah, I'm sorry about the confusion. The ratio test didn't work ): Glad you've got the answer now (:

 

[Dick]

Does following equation help you?

\frac{1}{2n}\leq \frac{1}{n^{1+1/n}}

That's an even better choice than the one I was thinking of.