# Limit proofs(indeterminate forms?)

• mynameisfunk
In summary: In fact, it never becomes true, as you can see by taking n = 5^{2k} for any integer k.You are supposed to be proving (a) and (b), not just one of them. Your proof should work for either one, but it doesn't.You are also not supposed to assume that a_n \geq 0. (a) and (b) might be true or false depending on the behavior of the a_n.(c) is not an indeterminate form; it is a statement. You are supposed to prove it.In summary, we are asked to determine the truth values of the following statements: (a) If the series \sum a_n^4 conver
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 $n \rightarrow \infty$.

## 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?

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?

mynameisfunk said:
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?

ystael said:
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 $\sum \frac{1}{n}$

mynameisfunk said:
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 $$n\to\infty$$.

mynameisfunk said:
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.

## What are indeterminate forms in limit proofs?

Indeterminate forms in limit proofs are mathematical expressions that cannot be solved using basic algebraic techniques. These forms involve variables that approach a particular value, but the outcome of the expression is undefined or cannot be determined.

## What are the most common types of indeterminate forms in limit proofs?

The most common types of indeterminate forms in limit proofs are 0/0, ∞/∞, 0∞, 1∞, and ∞-∞. These forms occur when both the numerator and denominator of a fraction approach 0 or infinity, or when a variable is raised to a power that approaches 0 or infinity.

## How do you solve indeterminate forms in limit proofs?

Indeterminate forms can be solved using techniques such as L'Hôpital's rule, factoring, and substitution. L'Hôpital's rule involves taking the derivative of the numerator and denominator separately and evaluating the limit again. Factoring can be used to simplify the expression and eliminate the indeterminate form. Substitution involves replacing the variable with a value that makes the expression solvable.

## What is the purpose of using indeterminate forms in limit proofs?

Indeterminate forms are used in limit proofs to evaluate the limit of a function at a point where it is undefined. These forms help determine the behavior of the function as it approaches a particular value and are essential in understanding the continuity and differentiability of a function.

## Are there any real-life applications of indeterminate forms in limit proofs?

Indeterminate forms in limit proofs have many real-life applications, such as in physics, engineering, and economics. For example, they can be used to calculate the velocity and acceleration of a moving object or to determine the optimal production level in a business. Indeterminate forms are also used in computer algorithms and data analysis.

Replies
1
Views
913
Replies
2
Views
1K
Replies
14
Views
1K
Replies
8
Views
1K
Replies
3
Views
2K
Replies
5
Views
796
Replies
3
Views
583
Replies
5
Views
1K
Replies
13
Views
2K
Replies
8
Views
2K