Limit proofs(indeterminate forms?)

[mynameisfunk]

limit proofs(indeterminate forms??)

Homework Statement

We work in the real numbers. Are the following true or false? Give a proof or counterexample.
(a) If \sum a^4_n converges, then \sum a^5_n converges.
(b) If \sum a^5_n converges, then \sum a^6_n converges.
(c) If a_n \geq 0 for all n, and \sum a_n converges, then na_n \rightarrow 0 as n \rightarrow \infty.
(d) If a_n \geq 0, for all n, and \sum a_n converges, then n(a_n - a_{n-1}) \rightarrow 0 as n \rightarrow \infty.
(e) If a_n is a decreasing sequence of positive numbers, and \sum a_n converges, then na_n \rightarrow 0 as n \rightarrow \infty.

Homework Equations

The Attempt at a Solution

(a) and (b) can be proved similarly. Since \sum a_n^4 converges, for some N, when n \geq N, then a_n^4 < 1. Take \beta s.t., a_n^4 < \beta < 1. That is, a_n < (\beta)^{1/4} < 1. Also, |(\beta)^{1/5}| < 1. This implies that |a_n^5| < 1 and therefore \sum |a_n^5| converges.

(c) This is where I get confused. THis seems like an indeterminate form, we have never done this in class. same for (d) and (e). any suggestions?

 

[ystael]

Your proof of (a) doesn't make sense. In order for a series \textstyle \sum b_n to converge, it is not enough that |b_n| < 1 for sufficiently large n!

Are you thinking of the root test? If so, there is certainly an exponent missing. But the converse of the root test is false: there are convergent series \textstyle\sum b_n having |b_n|^{1/n} \to 1 as n\to\infty. So you can't conclude from the convergence of \textstyle\sum a_n^4 anything about the behavior of |a_n^4|^{1/n}. The root test is not useful here.

For (a) and (b) you should be thinking about comparisons. Bear in mind that you are not given that a_n \geq 0, so the proof that works for one of (a) and (b) will not necessarily work for the other.

For (c)-(e): The condition na_n \to 0 should make you think of a particular series whose behavior you know. Try to modify this series to construct examples that test the boundaries of the conditions here.

 

[mynameisfunk]

How does a series converge if its terms do not converge to 0?

 

[Mark44]

How does a series converge if its terms do not converge to 0?

It doesn't. This is pretty much what the nth term test for divergence tells us. See http://en.wikipedia.org/wiki/Term_test.

 

[mynameisfunk]

so does my answer work? or no?

 

[mynameisfunk]

Are you thinking of the root test? If so, there is certainly an exponent missing. But the converse of the root test is false: there are convergent series \textstyle\sum b_n having |b_n|^{1/n} \to 1 as n\to\infty. So you can't conclude from the convergence of \textstyle\sum a_n^4 anything about the behavior of |a_n^4|^{1/n}. The root test is not useful here.

I think you might be mistaken here. Did you mean that there are sequences that where a_n \rightarrow 0 but \sum a_n doesn't converge? such as \sum \frac{1}{n}

 

[ystael]

I think you might be mistaken here. Did you mean that there are sequences that where a_n \rightarrow 0 but \sum a_n doesn't converge? such as \sum \frac{1}{n}

No, I meant precisely what I said: there are convergent series whose convergence is not detected by the root test, so the converse of the root test is false. \sum\frac1{n^2} is one; it converges by comparison to the telescoping series \sum\frac1{n(n+1)}, but |n^{-2}|^{1/n} \to 1 as n\to\infty.

 

[ystael]

so does my answer work? or no?

No. As I said, your argument for (a) establishes only that |a_n^5| < 1 for all sufficiently large n, which is very far from sufficient for convergence.