Prove that limit of $\sqrt{\frac{n^2+3}{2n+1}}$ is $\infty$ as $n$ tends to infinity.

Question

I need to prove that $\lim_\limits{n\to \infty}$$\sqrt{\frac{n^2+3}{2n+1}} = \infty$ (series) by using the definition:

"A sequence $a_n$ converges to $\infty$ if, for every number $M$, there exists $N∈N$ such that whenever $n≥N$ it follows that $a_n>M$."

I came with a proof which I'm not sure is valid, because I just learned the definition and maybe I didn't grasp it yet.

Please let me know what do you think of the following proof:

I'll find an $N$ such that whenever $n≥N$ it follows that $a_n>M$ for every $M$:

$M < \sqrt{\frac{n^2+3}{2n+1}} \Rightarrow M^2 < \frac{n^2+3}{2n+1} < \frac{n^2+n^2}{2n} = \frac{2n^2}{2n} = n$

Hence, we got $n > M^2$, therefore we can say that when $N = M^2$ then $a_n>M$ for every $n > M$. End of proof.

Do you think it is valid or am I missing something? Any suggestion for a better proof using the definition?

Thanks!

Answer 1

The proof isn't right. What you proved is that

$\frac{n^2+3}{2n+1}>N \implies n>M^2$ for $N=M^2$

This strategy would have worked if you had double implication type deductions... so then your reasoning would work backwards as well as forwards... but your implications are one way.

What you want to prove is that

$n>N \implies \frac{n^2+3}{2n+1}>M^2$, for some $N$

Also $n^2+3<2n^2$ requires $n \ge 2$ in the integers.

A way to check your proof is wrong is to try to go from $n>M^2$ to $\frac{n^2+3}{2n+1}>M^2$

You should do this at the end anyway. That is really your proof. The rest of the work is preliminary.

What you want to do is find an expression which is some constant times n which is less than $\frac{n^2+3}{2n+1}$ and force that expression to be greater than $M^2$

First I'm going to assume $n \ge 1$,

So then

$\frac{n^2+3}{2n+1}>\frac{n^2}{2n+1} \ge \frac{n^2}{2n+n} =\frac{n}{3}$ call this eqn.(1)

So given an $M$ we want $\frac{n}{3}>M^2$ so we want $n > 3M^2$

So here's the proof,

given an $M>0$, we choose $N = \max \{1,3M^2+1\}$

$\begin{aligned}&n \ge N \\ \\ \implies &n \ge 3M^2+1 > 3M^2 \\ \\ \implies &\frac{n}{3} > M^2 \end{aligned}$

Since we know $n \ge 1$ we can use eqn. (1), so we know that

$\begin{aligned} &\frac{n^2+3}{2n+1}>M^2 \\ \\ \implies &\sqrt{\frac{n^2+3}{2n+1}}>M \\ \\ \implies &a_n>M\end{aligned}$

EDIT: Note that I incorrectly had "min" before and fixed it to "max".

Answer 2

Your proof is mostly okay but you do three things wrong. One is linguistically wrong. You say, and quote,

"I'll find an $N$ such that whenever $n≥N$ it follows that $a_n>M$ for every $M$"

As stated, no such $N$ exists or can exist. $a_n >M$ for some $M$ but not for every $M$. For any $a_n$ we can always have an $M = a_n + 1 > a_n$ so there can't be any $a_n> M$ for all $M$.

The actually thing you wish to prove is that for any $M$ that we can choose as an arbitrary test limit we can find a point $N$ where suddenly, once we look at the values $a_n$ after $n=N$ we find they are all bigger than that $M$ we choose.

For example if we say $\lim_{n\to \infty} 2n = \infty$ what does that even mean? It means if we want to find a point where are the $a_n = 2n$ are bigger than $1000$ we can by looking at all $a_n$ where $n > 500$. In those cases $a_n > 1000$. But $a_{735} > 1000$ but we don't have $a_{735} > 1,000,000$. But if we want to find a point where all $a_n > 1,000,000$ we can do that by looking at all $a_n$ where $n > 500,000$. In those cases $a_n > 1,000,000$. But we don't have $a_{678,923} > 10^{100}$. Bue if we want to find a point where all $a_n > 10^{100}$ we just need $n > M = 5\cdot 10^{99}$. Then if $n > N$ then $a_n > 10^{100}$.

But we will never have a $N$ where $n>N$ means $a_n > M$ for all $M$. The best we can do, and it is a heck of a lot and it is all we actually want. Is that for any target we want, we can always find an $N$ for that target.

If $M$ is an number as big as we want, it could be $10^{(10^{100000}}$... or it could be $39$, but we can't pick an $M$ that is actually "infinite". But if $M$ is "as big as we like", then if $N = \frac M2$ thenfor every $n>\frac M2$ we will find that all $a_{n; n>N} = 2n > M$. That means whatever limit we want, we can always find a point where all $a_n$ past that point will all be bigger than that limit.

I'm picky that you understand what that means.

... okay... of my high horse

Secondly, you have you implication backwards.

You prove that if $a_n > M$ then $n > M^2$ but you didn't prove they other way around, if $n > M^2$ then $a_n > M$.

Thirdly you found an upper limit rather than a lower limit.

We want

$\sqrt{\frac {n^2 + 3}{2n + 1}} > M$.

And if we find anything $K > M$ we have to have $\sqrt{\frac {n^2 + 3}{2n + 1}}\ge K > M$. If we have $K > \sqrt{\frac {n^2 + 3}{2n + 1}}$ and $K > M$ that won't help us at all! (Consider we want to prove that $5 > 7$ and we get that $2*4 > 5$ and $2*4 > 7$ and therefore $5 > 7$ .... well, that's just wrong.)

$\sqrt {\frac {n^2 +3}{2n+1}} > M$ means

$\frac {n^2 + 3}{2n+1} > M^2$ (assuming $M > 0$). Now we must keep in mind this is what we WANT to show. THis is not what we actually know. We CAN'T conclude $\frac {n^2 + 3}{2n+1} > M^2\implies n >\frac {n^2 + 3}{2n+1} > M^2$ because we don't KNOW that $\frac {n^2 + 3}{2n+1} > M^2$.

we need to work the other way around $\frac {n^2 + 3}{2n+1} > M^2\Leftarrow ... something ...$.

And we have $\frac {n^2 + 3}{2n+1} > \frac {n^2}{2n+1} > \frac {n^2}{2n+n} = \frac n3$ (assuming $n\ge 1$)

And $\frac n3 > M^2 \Leftarrow$ (is implied BY) $n > 3M^2$.

So if $n > \max(1,3M^2)$ then $\frac {n^2 + 3}{2n+1} > \frac {n^2}{3n} = \frac n3 > M^2$ and so $a_n =\sqrt{\frac {n^2+3}{2n+1}} > |M| \ge M$.

And we are done.