# Convergent or divergent?

1. Feb 22, 2008

### 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?

2. Feb 22, 2008

### 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 $$\frac{1}{n^{1+\frac{1}{n}}}$$

3. Feb 22, 2008

### 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.

4. Feb 22, 2008

### 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.

Last edited: Feb 22, 2008
5. Feb 22, 2008

### 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)

6. Feb 22, 2008

### jostpuur

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.

7. Feb 22, 2008

### sutupidmath

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.

8. Feb 22, 2008

### sutupidmath

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

9. Feb 22, 2008

### 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.

Last edited: Feb 22, 2008
10. Feb 22, 2008

### 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.

Last edited: Feb 22, 2008
11. Feb 22, 2008

### 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!

12. Feb 22, 2008

### jostpuur

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

$$\sum_{n=1}^{\infty}\frac{1}{n^p}$$

converges if $p>1$. It follows, that if $a_n$ is a sequence so that $a_n>p>1$, then

$$\sum_{n=1}^{\infty} \frac{1}{n^{a_n}}$$

converges too, because

$$\frac{1}{n^{a_n}} \leq \frac{1}{n^p}\quad\quad\forall n$$

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

$$\sum_{n=1}^{\infty}\frac{1}{n^{a_n}}$$

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.

13. Feb 22, 2008

### sutupidmath

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