The sequence 1/n not convergent?

[xAxis]

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

 

[Dickfore]

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

 

[xAxis]

Ok, thanks. But I am 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.

 

[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 {a_n|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.

 

[xAxis]

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

 

[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.

 

[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.

 

[Skrew]

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.

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.

 

[Dickfore]

But, we're talking about a sequence, not a series. Besides, for the series, look at the partial sum:
<br /> 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}<br />

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