# Convergence of Series

1. Apr 29, 2012

### sid9221

$$\sum (1-\frac{1}{r})^{r^2}$$

Does this converge or diverge.(r=1..inf)

I have tried the following but do not think it is adequate(or correct for that matter)

$$(1-\frac{1}{r})^r (1-\frac{1}{r})^r = (1-\frac{1}{r})^{r^2}$$

and $$lim (1-\frac{1}{r})^r -> \frac{1}{e}$$

thats given from a previous part of this question.

So
$$lim (1-\frac{1}{r})^r (1-\frac{1}{r})^r -> \frac{1}{e^2}$$

Hence converges ?(As the limit exists)

Last edited: Apr 29, 2012
2. Apr 29, 2012

### sharks

According to the nth-term test for divergence, the limit=1, hence the series diverges.

P.S. Is that problem part of a question? If earlier sections of the problem are related to this question, you should post the whole thing.

3. Apr 29, 2012

### sid9221

Well its part of a group of questions but seemingly unrelated.

I've checked on wolfram that this series converges, so I don't think the test you quoted maybe valid...!?!

Wolfram say's to do a ratio test but that's not feasible by hand(at least to me)

4. Apr 29, 2012

### vela

Staff Emeritus
First, you have an error in your algebra. $x^n x^n = x^{2n}$ not $x^{n^2}$.

Second, you're confusing the convergence of the series $\sum a_n$ with the convergence of the sequence $a_n$. Just because the sequence converges doesn't mean the series converges.

Finally, what must $a_n$ converge to if the series is to converge?

5. Apr 29, 2012

### sid9221

Don't know what I was thinking !

So for it to converge the limit has to go to zero ? (Don't think the contrapostive of the non-null test is true though ?)

Say I put $$\sum[({1-\frac{1}{r}})^r]^2$$

Than took the limit of the inner part would I get $$\frac{1}{e^2}$$ It still would not be equal to zero...?

Last edited: Apr 29, 2012
6. Apr 29, 2012

### sharks

I'm not sure that i follow this. Is it based on any particular test? How do you arrive at that conclusion? I'm sorry for asking, but it seems that i messed up my understanding of sequences v/s series. I can relate this to AST, where the limit of $a_n$ needs to be zero for the series to converge.

7. Apr 29, 2012

### vela

Staff Emeritus
The contrapositive has to be true. It's logically equivalent to the original statement. The nth-term test says if $\displaystyle \lim_{n\to\infty}a_n\ne 0$ then the infinite series $\displaystyle \sum_n a_n$ will not converge. The contrapositive would be: if the series converges, then the limit of an has to be 0.

What I think you're thinking is if an goes to 0, it doesn't necessarily mean that the series converges. That is correct.

Right, so...

You've forgotten about the nth-term test. It's probably the very first test mentioned when you began to study series.

8. Apr 29, 2012

### sharks

I know about the nth-term test but its definition says that as long as the limit of the sequence does not equal zero, the series has to diverge. This theorem says nothing about convergence. If the limit is equal to zero, then according to the nth-term test, the series could either converge or diverge.

I'm not sure this is applicable to the nth-term test for divergence.

9. Apr 29, 2012

### vela

Staff Emeritus

10. Apr 29, 2012

### sharks

OK, you simply started with the end-result (that the series converges) and then backtracked to deduce that the limit has to be equal to zero.

11. Apr 29, 2012

### sid9221

I'm getting confused as well, an obvious example would be:

$$a_n = \frac{1}{n} -> 0$$ as n->infinity

But

$$\sum \frac{1}{n}$$

diverges ?

So how is the contrapositive true.

If your saying that we already know that the series is converging than that statement is true, that maybe different but in this case we don't know if it converges or diverges.