Sequences - Definition of convergence

[tolove]

Alright, I need some help with this.

a_n = \frac{1 - 5n^{4}}{n^{4} + 8n^{3}}

To find the limit of convergence, use l'Hopital's Rule. The result will come out to

L = -5

From my book,
"The sequence {a_n} converges to the number L if for every positive number ε there corresponds an integer N such that for all n,
n > N → | a_n - L | < ε"

So, to check that L = -5 is true, substitute in? How do I show that L = -5 using this definition?

| a_n - L | < ε

| \frac{1 - 5n^{4}}{n^{4} + 8n^{3}} - (-5) | < ε

Let n = 1, ε = 1

| \frac{1 - 5}{1 + 8} + \frac{45}{9}) | < ε = 1

\frac{41}{9} < ε = 1, which is not true

Thank you for your time! This definition is very confusing to me for some reason.

 

[WannabeNewton]

That's not how the definition works. You have to take an arbitrary $\epsilon > 0$ and show that there exists an $N \in \mathbb{N}$ (which may or may not depend on $\epsilon$) such that for all $n> N$ we have that $\left | \frac{1 - 5n^{4}}{n^{4} + 8n^{3}} + 5 \right | < \epsilon$.

 

[mathman]

Alright, I need some help with this.

a_n = \frac{1 - 5n^{4}}{n^{4} + 8n^{3}}

To find the limit of convergence, use l'Hopital's Rule. The result will come out to

L = -5

From my book,
"The sequence {a_n} converges to the number L if for every positive number ε there corresponds an integer N such that for all n,
n > N → | a_n - L | < ε"

So, to check that L = -5 is true, substitute in? How do I show that L = -5 using this definition?

| a_n - L | < ε

| \frac{1 - 5n^{4}}{n^{4} + 8n^{3}} - (-5) | < ε

Let n = 1, ε = 1

| \frac{1 - 5}{1 + 8} + \frac{45}{9}) | < ε = 1

\frac{41}{9} < ε = 1, which is not true

Thank you for your time! This definition is very confusing to me for some reason.

The basic idea is that for there is an N (large enough) so that the ε test will hold for all n > N.

 

[tolove]

Ohhh, so my e can be any positive number. For convergence, I must find an N.

thank you!

 

[WannabeNewton]

It might help to see a trivial example first. Consider the sequence given by $a_n = \frac{1}{n}$ and let $\epsilon > 0$. Intuitively we can see that this converges to $L = 0$ but let's show that this is indeed the case. There exists an $N\in \mathbb{N}$ such that $\epsilon N > 1$ i.e. $\frac{1}{N} < \epsilon$. Thus we have that $\left | \frac{1}{n} \right | < \epsilon$ for all $n\geq N$ meaning $a_n\rightarrow 0$, as we would expect.