the limit of this series through the definition

Question

Show that $$\lim_{n\to\infty}\frac{n^2+1}{n+2017}=\infty.$$

I start like that, for every $M>0$ there is an $N>0$, so that for every $n> N$, $$\frac{n^2+1}{n+2017}>M.$$

so how to continue these(according to the definition)? like without any advanced method like the Squeeze theorem and so on?

Answer 1

An option:

Let $n >2017.$

Then

$\dfrac{n}{2} = \dfrac{n^2}{2n} \lt \dfrac{n^2+1}{n+2017}.$

It suffices to show that

$\lim_{n \rightarrow \infty} (n/2)= \infty.$

Let $M$, real, positive be given.

Archimedes principle:

There is a $N$, positive integer, such that

$N > 2M.$

For $n \ge N$ we have

$n/2 \ge N/2 >M$.

Answer 2

As an alternative method, we have $$\small\lim_{n\to\infty}\frac{n^2+1}{n+2017}=\lim_{n\to\infty}\frac{(n+2017)^2-2\cdot2017n-2017^2+1}{n+2017}=\lim_{n\to\infty}\left[n+2017-\frac{2\cdot2017n+2017^2-1}{n+2017}\right]$$ and since $$\lim_{n\to\infty}\frac{2\cdot2017n+2017^2-1}{n+2017}=\lim_{n\to\infty}\frac{2\cdot2017+(2017^2-1)\cdot\frac1n}{1+2017\cdot\frac1n}=\frac{2\cdot2017}1=2\cdot2017$$ using $\lim_{n\to\infty}\frac1n=0$ along with the Combination Rules for limits, we have that $$\lim_{n\to\infty}\frac{n^2+1}{n+2017}=\left[\lim_{n\to\infty}n\right]+2017-2\cdot2017=\left[\lim_{n\to\infty}n\right]-2017=\infty$$

Answer 3

We have that

$$\frac{n^2+1}{n+2017}>M \iff n^2-Mn-2016>0$$

which by quadratic formula requires

$$n>\frac{M+\sqrt{M^2+8064}}{2}=k$$

thus it suffices assume $N=\lceil k \rceil $.