What's the value of $\lim_{n\to \infty} \frac{n+1}{n^2+1}\left[\frac{n^2+1}{n+1}\right]$

Question

Excuse my bad typing of mathematical symbols, I don't know how to type symbols. The $[ ]$ brackets are for the floor function where $[3.2]=3$ and $[2.9]=2$.

I have trouble calculating this limit, I know that floor function is mostly calculated by seperating it to two cases, where n is odd or even, but I can't go any further with this one. Any tips?

Answer 1

For the term in the floor function:

$$\left[\frac{n^2+1}{n+1}\right]=\left[\frac{n^2-1+2}{n+1}\right]=n-1+\left[\frac{2}{n+1}\right]$$

For sufficiently large $n$, $\left[\dfrac{2}{n+1}\right]=0$.

Answer 2

We have that

$$\frac{n^2+1}{n+1} = \frac{n^2+n-n+1}{n+1}=n-\frac{n-1}{n+1} \implies \left[\frac{n^2+1}{n+1}\right]=n-1$$

Answer 3

I would have approached the problem differently, abstaining from most math in favor of intuition. Let $\lfloor A \rfloor$ denote the floor of $A$.

It seems immediate to me that as $A \to \infty$ the ratio
$[E_1] \frac{\lfloor A\rfloor}{A}$ must go to $1$.

Edit per subsequent comment
$E_1$ holds, since $A-1 < \lfloor A \rfloor \leq A.$
Therefore, $\frac{A-1}{A} < \frac{\lfloor A\rfloor}{A} \leq \frac{A}{A} = 1.$

As $n \to \infty,$ the value
$\frac{n^2 + 1}{n+1}$ is clearly going to $\infty.$

Edit per subsequent comment
Therefore, based on $E_1$, as $n \to \infty$, $\displaystyle \frac{\lfloor \frac{n^2 + 1}{n+1} \rfloor}{\frac{n^2 + 1}{n+1}}$ is clearly going to $1$.

Since the above fraction is going to $1$, as $n \to \infty$, the numerator of the above fraction can be replaced by the denominator.

Once this replacement is made, the expression in the title of the OP's query exactly equals 1. Note that the OP's query is concerned about evaluating the limit as $n \to \infty.$ By the above analysis, when $n \to \infty,$ the corresponding replacement is valid.

Answer 4

We can use squeeze theorem too

$$ \lim_{n\to\infty}{\frac{n+1}{n^{2}+1}\left(\frac{n^{2}+1}{n+1}-1\right)}\leq\lim_{n\to\infty}{\frac{n+1}{n^{2}+1}\left[\frac{n^{2}+1}{n+1}\right]}\leq\lim_{n\to\infty}{\frac{n+1}{n^{2}+1}\frac{n^{2}+1}{n+1}} $$

The leftmost and rightmost terms approach $1$ as $n$ goes to infinity, then the middle term does too