# The sequence 1/n not convergent?

1. May 20, 2012

### xAxis

I read in Rudin's Analysis that sequence 1/n failes to converge in the set of positive real numbers. How comes?

2. May 20, 2012

### Dickfore

Because zero, the limit of the sequence, is not positive, but all of its elements are.

3. May 20, 2012

### xAxis

Ok, thanks. But Im a bit confused with some of his definitions. He calls the set n={1,2,3,...} the range of the sequence, and then says that the sequence is bounded if its range is bounded. Is this standard definition? I'm not sure if I've chosen the right book to brush on my maths.

4. May 21, 2012

### laughingebony

That's not standard. I don't have access to a copy of Rudin at the moment, but I suspect you've misread something. {1,2,3,...} (i.e., the set of natural numbers) is the domain of the sequence. The range is {an|n is a natural number} (i.e., the set of sequence values). With the correct definition of the range of a sequence, the definition you gave of a bounded sequence is standard.

5. May 21, 2012

### xAxis

Yes, you are right. I missread. Everything is fine and this is an excellent book. Thank you.

6. May 22, 2012

### Noesis

Note that to say the sequence is bounded if the range is bounded is the same thing in this case. Nothing new beyond definitions is said here. However, interesting and new things regarding the convergence of a sequence can be said if the sequence is bounded.

The sequence/range is bounded if there is some number that is greater than all elements contained within the sequence/range.

7. May 23, 2012

### HallsofIvy

There is a theorem (the "r test") that says that $$\sum n^{r}$$ converges if and only if r< 1 (or that $\sum 1/n^r$ converges if and only if r> 1.)

So "$\sum 1/n$" is a borderline case: it diverges by the integral test:
$\int_1^\infty dx/x$ does not converge so the series does not converge.

Last edited by a moderator: May 23, 2012
8. May 24, 2012

### Skrew

You might run into a circular argument depending upon how you have defined/proved the properties of ln(x). The proof that ln(x) ->00 may or may not depend upon 1/n diverging to infinity.

9. May 24, 2012

### Dickfore

But, we're talking about a sequence, not a series. Besides, for the series, look at the partial sum:
$$S_{2^{n}} = \frac{1}{1} + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \ldots + \left( \frac{1}{2^{n - 1} + 1} + \ldots + \frac{1}{2^{n}} \right) \ge 1 + \frac{1}{2} + \frac{2}{4} + \ldots + \frac{2^{n - 1}}{2^{n}} = \frac{n + 1}{2}$$

Since the sequence on the right is unbounded, so is the subsequence of partial sums on the left. Hence, the series is divergent.