Directly prove using the definition of convergence $\lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} = \infty$

Question

The definition of convergence is : $\lim\limits_{n \to \infty}a_n =l \Rightarrow \lim\limits_{n \to \infty}a_n -l=0 $

So $\lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} =\infty \Rightarrow \lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} -\infty =0$

and we have to show that $ \forall n > X,|\frac{n^2 +1 }{4n -1} -\infty| < \epsilon $.

Answer 1

$$4n-1\leq 4n\quad\text{and}\quad n^2+1\geq n^2.$$ Therefore $$\frac{n^2+1}{4n-1}\geq \frac{n^2}{4n}=\frac{n}{4}.$$ Then, if $M>0$, if you take $N=\lfloor 4M\rfloor+1$, you have that $$\frac{n^2+1}{4n-1}>M$$ when $n\geq N$ what prove the claim.

Answer 2

To show that a function f(n) goes to $\infty$ as n goes to $\infty$, we need to show that for any $\alpha$ > 0, there will be a $\delta$ > 0 such that for all n > $\delta$, f(n) > $\alpha$.

You function is f(n) = $\frac{n^2 +1 }{4n -1}$ >$\frac{n^2 }{4n}$ > $\frac{n}{4}$

and we need the f(n) > $\alpha$, which can be satisfied if we have $\frac{n}{4}$ > $\alpha$ i.e. n > 4$\alpha$. So our required $\delta$ = 4$\alpha$ and you have shown the required thing.

Answer 3

It's an old question but as I get pretty the same one I choosed to answer it fully so maybe it will help other futur student and mostly it helps me to improve myself on rigourous mathematical writting.

First let's recall that the true definition of $\lim_n a_n = \infty $ is: $\forall M \in \mathbb{R}^+ \; \exists N \in \mathbb{N} \; s.t. \; \forall n > N \Rightarrow a_n>M $

First let pay attention to the following inequalities:

• $n^2+1>n^2$ as $1>0$
• $4n-1<4n$ as $-1<0$
• So by the two inequalities above: $\frac{n^2+1}{4n-1}>\frac{n^2}{4n}=\frac{n}{4}$
• So now for all $M>0$ if $N=4M \Rightarrow \forall n>N=4M$ we get the following inequality $\frac{n}{4}>\frac{N}{4}=\frac{4M}{4}=M$

We can conclude by writting the above definition with the help of all the inequalities we ve written above.
$\forall M \in \mathbb{R}^+ \; \exists N=4M \in \mathbb{N} \; s.t. \; \forall n > N=4M \Rightarrow a_n=\frac{n^2+1}{4n-1}>\frac{n^2}{4n}=\frac{n}{4}>\frac{N}{4}=\frac{4M}{4}=M$
Which is your proof by definition.