Using definition of a limit to prove that $\lim_{n \rightarrow \infty} \frac{2}{n^4} = 0$

Question

Using the definition of a limit, prove that $$\lim_{n \rightarrow \infty} \frac{2}{n^4} = 0$$

My take: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$

$$\left |\frac{2}{n^4} - 0 \right | < \epsilon$$

Notice that $\frac{2}{n^4} < \frac{2}{n} < \epsilon$, so I think I can say $n > 2\epsilon$ (not sure if I can do this).

So, given all that rough work,

Proof: Given $\epsilon >0$, choose $N > 2\epsilon$ and suppose $n \ge N$ such that

$$\left | \frac{2}{n^4} \right | < \frac{2}{n} \le \frac{2}{N} \le \frac{1}{(2\epsilon)^2} = \epsilon?$$

I'm sure I screwed it up somewhere. Can someone help me please?

Answer 1

I continue on your proof.

Your take: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$

$$\left |\frac{2}{n^4} - 0 \right | < \epsilon$$

Let $\epsilon>0$, we want to find an $N$ such that $\forall n\in\mathbb{N},~ n\ge N\Longrightarrow\left |\frac{2}{n^4}\right | < \epsilon$.

Observe on this inequality: $\left |\frac{2}{n^4}\right | < \epsilon$. Since the "$n$" we discuss is only under natural numbers, so we roughly assume that $n\in\mathbb{N}$. Then we can go on our observations:

$$\begin{alignat}{2} &\left |\frac{2}{n^4}\right | < \epsilon\\ \underset{~n\in\mathbb{N}}{\Longleftrightarrow}&\frac{2}{n^4}<\epsilon\\ \Longleftarrow &~\frac{2}{n^4}<\underbrace{\frac{2}{n}<\epsilon}_{\text{hope it can be true}} \end{alignat} $$

Hence, from the "$\Longleftarrow$", when $\frac{2}{n}<\epsilon$, then $\left |\frac{2}{n^4}\right | < \epsilon$ can be true. But when will $\frac{2}{n}<\epsilon$? $$\begin{alignat}{2} &\frac{2}{n}<\epsilon\\ \Longleftrightarrow&2<n\cdot\epsilon\\ \Longleftrightarrow&n>\frac{2}{\epsilon} \end{alignat} $$

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So until now, we have gain that $\displaystyle\forall n\in\mathbb{N},~n>\frac{2}{\epsilon}\Longrightarrow\left|\frac{2}{n^4} - 0 \right| < \epsilon$. Then we choose $N$ to be $\lceil\frac{2}{\epsilon}\rceil$, then $\forall n\geq N,~\left|\frac{2}{n^4} - 0 \right| < \epsilon$. $\square$

Answer 2

Choose $N>2/\epsilon$. Then $|2/N^4|<2/N<\epsilon$.

In general when solving limit problems, you ought to think of $\epsilon$ as being very small, and $N$ being large. The smaller $\epsilon$ is, the larger $N$ must be. So a proper relationship between $N$ and $\epsilon$ will usually indicate that $N$ increases as $\epsilon$ decreases.

Answer 3

There is a more elegant solution using the an archimedian property of the real numbers; and, also, these problems are easy if you remember that the general technique for any limit definition problem is to either "clean up" the absolute value in the limit definition using a combination of algebra and knowledge of absolute value inequalities, or to try to bound the absolute value term with another term that's easier to work with; and possibly some combination of all of the above, combined with some random theorem you previously encountered the course.

I'll supply the proof with a commentary to illustrate the proper thought process:

Let's look at what you want to prove: $$(1) \forall \epsilon>0, \exists m\in\mathbb N \ni (n \in N\land n>m \rightarrow \left | \frac {2}{n^4}-0 \right |<\epsilon). $$

I am now planning an approach in my mind that uses the algebraic/bounding/absolute value inequality techniques mentioned in the beginning paragraph. Mentally, I glance back at every thing I've learned so far in the entire course and pause at the 3 archimedian properties found in the suprema unit of all textbooks on introductory real analysis. What catches my attention is the $\frac {1}{n}$ part of the logical formula for the archimedian property $\forall\epsilon>0,\exists n\in\mathbb N \ni 0<\frac{1}{n}<\epsilon$

This $\frac {1}{n}$ term reminds me a lot of the $\frac{2}{n^4}$ term in the absolute value inequality $\left | \frac {2}{n^4}-0\right |<\epsilon.$ In fact, the subformula $\left | \frac {2}{n^4}-0\right |<\epsilon$, in the limit definition given by $\forall \epsilon>0, \exists m\in\mathbb N \ni (n \in N\land n>m \rightarrow \left | \frac {2}{n^4}-0 \right |<\epsilon)$ cleans up very nicely in this problem to obtain

$$(2) \forall \epsilon>0, \exists m\in\mathbb N \ni (n \in N\land n>m \rightarrow \frac {2}{n^4}<\epsilon)\textrm{, since } 2/n^4=\left | \frac {2}{n^4}-0\right | \textrm{ because } n^4>0$$

You need to, and should, be able to verify in your head that (1) and (2) are logically equivalent (i.e. there is a double implication connecting them logically) and realize that it is significantly easier to prove (2) than it is to prove 1; and also know that proving (2) is sufficient for this problem because (2) will imply (1), which is what we need to prove. I now attempt to do so:

It seems as if I could use the axioms of $\mathbb R$ to "build" $\frac {2}{n^4}<\epsilon,$ starting with $\frac{1}{n}<\frac {\epsilon}{\sqrt[4]{2}}$ using the archimedian property described above on $\frac {\epsilon}{\sqrt[4]{2}} \textrm{in place of } \epsilon$ Let me demonstrate: $$\textrm{"Let }\epsilon>0\textrm{and consider }\sqrt[4]{\frac {\epsilon}{2}}.$$ $$\textrm{Since}, \sqrt[4]{\frac {\epsilon}{2}}>0, \textrm{the archimedian property stated above }\textrm{guarantees }\exists m\in\mathbb N\ni 0<\frac{1}{m}<\sqrt[4]{\frac {\epsilon}{2}}.$$$$\textrm{Let } n\in\mathbb N \land m<n.$$$$ \textrm{ This is where we begin to build the desired inequality: }$$$$\textrm{Firstly, since 0<m<n, we conclude that }0<\frac{1}{n}<\frac{1}{m}.$$$$\textrm{Secondly, since } 0<\frac{1}{n}<\frac{1}{m}\land \frac{1}{m}<\sqrt[4]{\frac {\epsilon}{2}}, \textrm{we conclude (using transitivity) that } $$$$ 0<\frac{1}{n}<\sqrt[4]{\frac {\epsilon}{2}}\textrm{ is true. }$$$$\textrm{By raising the inequality } 0<\frac{1}{n}<\sqrt[4]{\frac {\epsilon}{2}}\textrm{, to the 4th power and multiplying by 2, I conclude that } \frac{2}{n^4}<\epsilon.$$$$\textrm{We have now proven the statement we labled as 2. That is, } \forall \epsilon>0,\exists m\in\mathbb N\ni (n\in\mathbb N\land m<n \rightarrow \frac {2}{n^4}<\epsilon. $$$$\textrm{As was already pointed out, this then sufficiently implies that the original statement, }$$$$ \forall \epsilon>0, \exists m\in\mathbb N\ni(n\in \mathbb N\land m<n \rightarrow \left |\frac {2}{n^4}-0\right|<\epsilon)\textrm { is true, and thus the proof is complete."}$$