Convergence of a sequence as n tends to infinity updated solution

Question

Hi I wrote a initial solution to the question I posed I will attach it here.

I want to do a proof for the convergence of a sequence, I attach my initial solution containing the question below:

I have two general questions if someone would be kind enough to answer them:

a) What is the meaning of the square bracket i.e. N := $\lceil \frac{3}{\epsilon} \rceil$

b) Is the general proof in its entireity correct.

This is the original question I posed: Convergence of a sequence as n tends to infinity

I was thinking about re-editing the old post, but was worried it may come to cluttered.

My main question is should the end of the second line in black ink be $\frac{5}{n} + \frac{2}{n^2}$ or is the negative sign right, I am in doubt due to the previous question I had from here: Convergence of a sequence solution mistake?

Thank you for any responses

Answer 1

There are distinct symbols here:

• $\lceil x\rceil$ means the smallest integer that is more than or equal to $x$. For instance, $\lceil 7.3 \rceil=8$

• $\lfloor x\rfloor$ means the largest integer that is less than or equal to $x$. For instance, $\lfloor 7.3\rfloor=7$

• I belive $[x]$ means the integer part of $x$ and $\{x\}$ the fractional part of $x$. Hence, $[7.3]=7$ and $\{7.3\}=0.3$. There isn't consensus on this notation for negative $x$ though, some use $[-2.3]=-2$ and some use $[-2.3]=-3$. However, it's always true that $[x]+\{x\}=x$.

With regards to the proof, it is needlessly convoluted I believe. Since you got to the point that

$$\frac{5}{n}-\frac{2}{n^2} <\frac{5}{n}$$

It suffices to show that given $\epsilon >0$ there is $m\in \mathbb{N}$ such that for all $n\geq m$, $\frac{5}{n}<\epsilon$. This is equivalent to $m>\frac5\epsilon$, which follows from the archimedean property.