Is My Thinking Wrong About Convergent or Divergent?

[Shenlong08]

For lim n->infinity n^-(1+1/n), the p series test shows that it converges since (1+1/n) will be greater than 1, while doing a limit comparison test with 1/n gives 1 showing that it diverges since 1/n diverges. For which one is my thinking wrong about?

 

[sutupidmath]

well you cannot actually compare this to the harmonic series 1/n, since

\frac{1}{n^{1+\frac{1}{n}}}= \frac{1}{n n^{\frac{1}{n}}} now since

n^{\frac{1}{n}}>= 1, you can prove this using induction!

we get:

\frac{1}{n^{1+\frac{1}{n}}}= \frac{1}{n n^{\frac{1}{n}}}=<\frac{1}{n}

so the divergence of 1/n tells you nothing about the series \frac{1}{n^{1+\frac{1}{n}}}

 

[Shenlong08]

I'm not comparing the sequence by the above/below method, I'm saying if I compare it to the harmonic series through a ratio, then

lim n-> infinity [ n*n^(1/n) ] / n

lim n-> infinity n^(1/n)

lim n-> infinity (1/n) ln(n)

lim n-> infinity 1/n ...l'hopitals'

= 0 -> e^0=1

and if n->infinity converges to a finite number > 0, then they must either both converge or both diverge.

 

[sutupidmath]

What is the derivative of 1/n when n=1,2,3,4,5,... ?
And i do not see at all how you are comparing it to the harmonic series, all i see is comparing to a series n.

 

[Shenlong08]

The derivative was taken from the line (1/n)ln(n). 1/n was evaluated as 0.

1/n / 1/[ n*n^(1/n) ] (sorry i don't know how to make it more clear)

 

[jostpuur]

For lim n->infinity n^-(1+1/n), the p series test shows that it converges since (1+1/n) will be greater than 1, while doing a limit comparison test with 1/n gives 1 showing that it diverges since 1/n diverges. For which one is my thinking wrong about?

I'm not fully sure what is the theorem you are using, but I have a guess about your mistake. It is not going to be sufficient to have 1+\frac{1}{n}>1 for convergence, you would need to have 1+\frac{1}{n}>p with some fixed p>1, which in this case is not true.

 

[sutupidmath]

I'm not fully sure what is the theorem you are using, but I have a guess about your mistake. It is not going to be sufficient to have 1+\frac{1}{n}>1 for convergence, you would need to have 1+\frac{1}{n}>p with some fixed p>1, which in this case is not true.

The p-series is given by
sigma 1/n^p = 1/1p + 1/2p + 1/3p + ...
where p > 0 by definition.
If p > 1, then the series converges.
If 0 < p <= 1 then the series diverges.

 

[sutupidmath]

I'm not fully sure what is the theorem you are using, but I have a guess about your mistake. It is not going to be sufficient to have 1+\frac{1}{n}&gt;1 for convergence, you would need to have 1+\frac{1}{n}&gt;p with some fixed p&gt;1, which in this case is not true.

Why wouldn't 1+1/n>1 be sufficient?

 

[d_leet]

How has this thread lasted so long?! By the limit comparison test with the harmonic series. we find this series clearly diverges, if you are not familiar with this test look in any calculus book in the sections about convergence of series.

Edit: This is covered in volume one of Apostol on page 396, this is the only calculus text I have readiy available, but I know this test will be covered in nearly all calculus texts.

 

[Shenlong08]

I merely wanted to know which test I was using wrong, if you care to explain why the p series test does not apply, it'd be appreciated.

jostpuur gave a reason, which if you verified, would be great.

 

[sutupidmath]

I think the p series test does not apply because it applies only on those cases where p-is merely a constant, that is the power is a constant, while on your problem you have on the power 1+1/n, so this changes also with the change on n, which means that it is not a constant!

 

[jostpuur]

sutupidmath is answering his own question now. It could be I can make it even clearer. So you know that

<br /> \sum_{n=1}^{\infty}\frac{1}{n^p}<br />

converges if p&gt;1. It follows, that if a_n is a sequence so that a_n&gt;p&gt;1, then

<br /> \sum_{n=1}^{\infty} \frac{1}{n^{a_n}}<br />

converges too, because

<br /> \frac{1}{n^{a_n}} \leq \frac{1}{n^p}\quad\quad\forall n<br />

Then suppose you have a sequence a_n so that a_n&gt;1 and \lim_{n\to\infty} a_n = 1. Can you explain why

<br /> \sum_{n=1}^{\infty}\frac{1}{n^{a_n}}<br />

would converge now? Can you put some convergent series above it? You cannot. So, there's no reason to believe why this series would converge. The example of the OP is a counter example, that proofs that series like this indeed do not necessarily converge.

 

[sutupidmath]

sutupidmath is answering his own question now.

Yeah, indeed i answered my own question! I was confused for a moment!