# Homework Help: Convergent sequence

1. Oct 31, 2015

### squenshl

1. The problem statement, all variables and given/known data
Show, from the definition of what it means for a function to converge to a limit, that the sequence $\left\{x^t\right\}_{t=1}^{\infty}$ with $x^t = \frac{2t+5}{t^2+7}$ converges to $0$ as $t$ goes to infinity.

2. Relevant equations
A sequence converges to $x^0 \in X$ if for any $\epsilon > 0$, there is $N \in \mathbb{N}$ such that if $t > N$, then $d(x^t,x^0) < \epsilon$.

3. The attempt at a solution
To show that $x^t = \frac{2t+5}{t^2+7}$ converges to $0$ we must, for any $\epsilon > 0$, find a value $N$ such that if $t > N$, then
$$\left|\frac{2t+5}{t^2+7} - 0\right| = \left|\frac{2t+5}{t^2+7}\right| < \epsilon$$.
Now sure how to simplify $x^t$ to show that it is less than or equal to some much simpler expression in $t$ that can clearly be made less than any given $\epsilon$ by choosing $n$ large enough. Please help.

2. Oct 31, 2015

### PeroK

First, you could note that $t^2 +7 > t^2$

3. Nov 1, 2015

### HallsofIvy

Since t is positive so is the fraction so you can drop the absolute value sign: $\frac{2t+ 5}{t^2+ 7}< \epsilon$ so $2t+ 5< \epsilon(t^2+ 7)$. For what t is $\epsilon t^2- 2t+ 7\epsilon- 5> 0$?

4. Nov 1, 2015

### Ray Vickson

If you do not want the smallest possible $N$ giving $|x_t| < \epsilon$ for $t > N$, but are satisfied with some $N = N(\epsilon)$ that 'works', the problem is quite straightforward. Do you agree that for $t \geq 1$ the numerator is $\leq 7t$? Do you agree that the denominator is $> t^2$?

5. Nov 1, 2015

### squenshl

Yup that's obvious so is it as easy as saying $\frac{2t}{t^2} = \frac{2}{t} < \epsilon$.

6. Nov 1, 2015

### Ray Vickson

Well, no: $5 + 2 \neq 2$.

7. Nov 1, 2015

### squenshl

Then I'm a little stuck. That $t^2$ has lost me. It's easy if it's just $t$.

8. Nov 1, 2015

### Ray Vickson

$\frac{7t}{t^2} = \frac{7}{t}$. Note that the original numerator is $7 t$, not $2 t$ as you wrote (but otherwise you did exactly the same simplification, using 2 instead of 2+5 = 7).

9. Nov 1, 2015

### squenshl

Ok why is the numerator $\leq 7t$ for $t > 0$??

10. Nov 1, 2015

### Ray Vickson

It isn't, and I never claimed it was. Go back and read what I wrote, word-for-word.

11. Nov 1, 2015

### squenshl

So is it as easy as saying $\frac{7t}{t^2} = \frac{7}{t} < \epsilon$??

Last edited: Nov 1, 2015
12. Nov 1, 2015

### krebs

EDIT:
Only if you understand how you got there...

Solve the two inequalities in terms of $t$:

$2t + 5 \leq 7t$

and

$\frac{1}{t^2 + 7} \leq \frac{1}{t^2}$

What relationship do you get for t in each one? What does $\left\{x^t\right\}_{t=1}^{\infty}$ dictate for t?

If $A \leq B$ and $B < C$, then $A < C$, yes?

Last edited: Nov 1, 2015
13. Nov 1, 2015

### squenshl

For the first one $t \geq 1$ and second one $t > 0$. Where am I going with this??

14. Nov 2, 2015

### squenshl

So $\frac{2t+5}{t^2+7} \leq \frac{7t}{t^2} = \frac{7}{t}$ for $t > 0$??

15. Nov 2, 2015

### krebs

Perfect! That's very good. (except it's for $t \geq 1$, look at post #13 again... And our sequence specifies valid values for t (also t > n, which is a natural number...). Is our inequality ever violated?)

And if we can show $\frac{7}{t} < \epsilon$, then we know that $\frac{2t+5}{t^2+7} < \epsilon$, right?

If $n \in \mathbb{N}$, is $(7 \times n) \in \mathbb{N}$, as well?
If $t > n$ does it hold that $7 \times t > 7 \times n$?

(Note: I will use lowecase n because I don't want to make it confusing between the variable and the set)

Last edited: Nov 2, 2015
16. Nov 2, 2015

### krebs

Sorry, you were wrong in #13 and I didn't catch it. It's a minor error, but

$\frac{1}{t^2 + 7} \leq \frac{1}{t^2}$

$t^2 \leq t^2+7$

$0 \leq 7$

thus, the second inequality is always true.

17. Nov 2, 2015

### PeroK

One thing you're missing is that you don't need inequalities to hold for all $t > 0$. If you are taking the limit $t \rightarrow \infty$ then you can look for inequalities that hold for large enough t. For example,

For $t > 5$ we have $2t + 5 < 3t$

Last edited: Nov 2, 2015
18. Nov 2, 2015

### squenshl

So $t\geq 1$.

19. Nov 2, 2015

### squenshl

Not sure what u mean here.

20. Nov 2, 2015

### squenshl

I chose $N > \frac{7}{\epsilon}$ and we shall have if that $t > N$, then $$\left|\frac{2t+5}{t^2+7}\right| < \epsilon.$$

21. Nov 2, 2015

### PeroK

What you are trying to do with this problem is show what happens to your function as $t \rightarrow \infty$. This means that when you are working with your function, you can assume that $t$ is greater than a given value. Any concerns about what happens if $t = 0$ or $t = 1$ are irrelevant to the limit. Let me given you an example.

Suppose I wanted to show that $t^3 + 6t^2 - 15t + 173 \rightarrow \infty$ as $t \rightarrow \infty$

Well, I could try to solve cubic equations. But, instead, I would do something like this:

For $t > 0, \ t^3 + 6t^2 - 15t + 173 > t^3 - 15t = t(t^2 - 15)$

For $t > 4, \ t(t^2 - 15) > t$

Hence, for $t > 4, \ t^3 + 6t^2 - 15t + 173 > t$

And, then it's very easy.

22. Nov 2, 2015

### squenshl

Ok thanks!!

23. Nov 2, 2015

### krebs

I think you might be caught up on the wrong things. We are working backwards along the following proof. Normally I wouldn't post an entire proof, but you've already solved the problem. I want to make sure you have a complete understanding of how that solution worked.

1. There exists a natural number $a$, such that $x > a$ satisfies the inequality: $\frac{1}{x} < \epsilon$ This is known/obviously true because $\lim_{x \to ∞} \frac{1}{x} = 0 < \epsilon$
2. $\frac{7}{7x} < \epsilon$ 1
3. $\{7, a\} \in ℕ$ 1 and definition of ℕ
4. $7a \in ℕ$ 3 and closure of ℕ for multiplication
5. let $n = 7a$
let $t = 7x$
6. $7x > 7a$ 1
$t > n$ 1,5 and 7x > 7a
7. $n \in ℕ$ 4,5
8. $\frac{7}{t} < \epsilon$ 2, 5
$\frac{7t}{t^2} < \epsilon$
To recap: We have shown that there exists a number "n" in ℕ, and that if t > n, then $\frac{7}{t} < \epsilon$. The rest if the proof is algebra to show that $\frac{5+2t}{t^2+7} < \frac{7}{t} < \epsilon$
9. $7 > 0$ axiom
$t^2 + 7 > t^2$
$\frac{1}{t^2} > \frac{1}{t^2+7}$
$\frac{7t}{t^2} > \frac{7t}{t^2+7}$
10. $t > 1$ This is a new constraint on our solution that t is greater than 1, but it doesn't matter because we can make n (and therefore t) arbitrarily large
$5t > 5$
$7t > 5+2t$
$\frac{7t}{t^2+7} > \frac{5+2t}{t^2+7}$
11. $\frac{5+2t}{t^2+7} < \frac{7t}{t^2+7} < \frac{7t}{t^2} < \epsilon$ 10, 9, 8

∴ $\frac{5+2t}{t^2+7} < \epsilon$ 11, and $t > n$ 6, and $n \in ℕ$ 7

Last edited: Nov 2, 2015
24. Nov 2, 2015

### Ray Vickson

You were the one that said $\{x_t\}_{t=1}^{\infty}$ in post # 1, so that means you want $t = 1, 2, 3, \ldots$. In other words, you only care about $t \geq 1$. But, in fact, to analyze what happens as $t \to \infty$ you could assume $t > 1,000,000$ if you wanted to---because eventually you will be taking $t$ larger than any number you can name or write down. That's what $t \to \infty$ means.

So, $t \geq 1$, agreed? Now, for $t \geq 1$ we have $5 \leq 5 t$, so $2t + 5 \leq 2t + 5t = 7t$. That's all there is to it: just basic properties of numbers.

You also have $t^2 + 7 > t^2$, so for $t \geq 1$ you have
$$0 < \frac{2t+5}{t^2+7} < \frac{7t}{t^2} = \frac{7}{t}$$

25. Nov 2, 2015

### geoffrey159

If you don't like the given sequence, you can work the problem on a simple equivalent in a neighborhood of $+\infty$.