# Limit proofs(indeterminate forms?)

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 $n \rightarrow \infty$.

## 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?

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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.

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

Mark44
Mentor

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.

so does my answer work? or no?

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

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 $$n\to\infty$$.

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.