# Convergence of a sequence with practical example and more complex question about the workings of a sequence

I am really trying to plug in a point to gain a practical understanding.

For $$N \in \mathbb{N}$$

$$x_n = 3(1 + \frac{1}{n}) \to 3$$ as $$n \to \infty$$

$$\vert 3 + \frac{3}{n} -3 \vert \lt \epsilon$$

$$\vert \frac{3}{n} \vert \lt \epsilon$$

since $$n$$ is always positive

$$\frac{3}{n} \lt \epsilon$$

There exists an $$N$$ such that $$\frac{3}{\epsilon} \lt N$$

Now I want pick a random number say $$\epsilon = .002$$

Then in order for $$x_n$$ to be closer to $$a$$ then $$\epsilon$$ we solve for $$N$$

$$\frac{3}{.002} \leq N$$

which is $$1,500 \leq N$$

lets choose $$1,600$$

$$3 + \frac{3}{1,600} = 3.001875$$

$$\vert 3.001875 - 3 \vert = .001875 \lt .002$$

Is this correct? Should $$N$$ be strictly greater than the fraction or is it acceptable to have have $$N \geq \frac{3}{\epsilon}$$ Also these seem like special cases where the algebra just happens to work out perfectly, what are you supposed to do if it doesn't? How do you even tell if it converges? If you can't isolate n from the sequence like in this example then is that a pretty good indicator that it may not converge or that you may not be able to find $$N \in \mathbb{N} S.T.......$$?

A big question of mine is what is the difference between lets say $$1 + 2x_n$$ and $$2-\frac{1}{n}$$ They have very different techniques in the book.

Yes, there is no mistake in your example, and actually it has sense. I read it as « in order to be close to 3, $$N$$ has to be large », especifically, larget than $$1500$$ if you want the error be smaller than $$0.002$$.

For your second question « what if you cannot isolate $$n$$? ». No, it is not an indicator of not convergence of the sequence. In my opinion, a good indicator might be to try to recognize other well known limits inside the limit your currently studying. So, if you only recognize well known comvergent sequences, it is much easier to figure out if the new sequence converges or not. If you recognize not convergent sequence, then you have to be careful because can be some cancelations which can make the new sequence a convergent one. Since you only have two possibilities (it converges or not), if you don’t recognize any well-known sequence and cannot prove the convergence either, you can try to find a counterexample to prove that the sequence does not converge, or to apply some theorem (like L’hôpital rule, super useful theorem to find limits of sequences) in order to prove the (not) converge.

Finally, for your last question, there is no difference between your two example. Note that the only interesting part of both sequence is the factor $$\dfrac{c}{n}$$, where $$c$$ is some constant depending on if you are in the first example or in the second one. Note that in both cases, independently pf what $$c$$ is, the limit $$\dfrac{c}{n}\to0$$ when $$n$$ goes to infinity. So, why they have so different techniques? Probably just because pedagogical purposes. I guess they wantsd to show two different methods to solve the same kind of problems, but I insist, both problem arr essentially the same (it only changes the constant $$c$$).

Quick check of the definition of the limit of a sequence: The sequence $$(x_n)$$ converges to $$x$$ if for all $$\epsilon > 0$$ there exists a non-negative integer $$N$$ such that if $$n \geq N$$ then $$|x_n - x| < \epsilon$$.

What you have there looks correct to me.

I tend to obtain $$N$$ so that the inequality for $$\epsilon$$ is satisfied strictly. I do this to match the definition exactly so that I know that I can use any results based on that definition.

If I have understood you correctly then you have raised two important points:

• Existence of the limit vs actually calculating it. Sometimes the value $$x$$ may not be apparent (follow the link above for examples).

• Existence of the limit vs calculating the rate of convergence. For the definition to work, we only need to know that an $$N$$ exists for any given $$\epsilon$$ in principle. We don’t necessarily need to solve for it explicitly, but an explicit solution is a very good way of establishing existence. Explicit calculation of $$N$$ for $$\epsilon$$ does tell us how big $$N$$ needs to be to achieve $$\epsilon$$ — the “rate of convergence”.

In general it can indeed be hard to tell whether an arbitrary sequence converges and to calculate the limit if it does. But there are increasingly sophisticated theorems that can give results (for example the Squeeze Theorem).