$\lim_{x \to \infty}\frac{x^{2(n+1)}}{(x^2+1)^{n+1}}$ and convergence of $\int_{0}^{\infty}\frac{dx}{(x^2+1)^{n+1}}$ I was told that in order to prove the convergence of 
$$\int_{0}^{\infty}\frac{dx}{(x^2+1)^{n+1}}$$
I shuold do the comparison test, the limit edition. 
And to take: 
$$
g(x) = \frac{1}{x^{2(n+1)}}
$$
So i tried, but i dont get how they got to a limit here: 
Lets take: 
$$
f(x) = \frac{dx}{(x^2+1)^{n+1}}
$$
Now lets check: 
$$
\lim_{x \to \infty}\frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{x^{2(n+1)}}{(x^2+1)^{n+1}} = ?
$$
I probably miss something basic, sorry if its a stupid question. 
Maybe i could somehow say something of the sense of: 
$$
\lim_{x \to \infty} (1+x) = \lim_{x \to \infty}(x)
$$
And then the limit is $1$?
Thank you and sorry for the question. 
 A: If you were to calculate $(x^2+1)^{n+1}$, you would find that the term of highest degree would be $(x^2)^{n+1} = x^{2(n+1)}$. In the infinite limit, the terms of lesser degree don't matter - intuitively, their growth is overwhelmed by the term of highest degree - so you can conclude
$$\lim_{x \to \infty}\frac{x^{2(n+1)}}{(x^2+1)^{n+1}} = \lim_{x \to \infty} \frac{x^{2(n+1)}}{x^{2(n+1)}}$$
for which the limit is obvious.

A simpler example of this fact can come by considering this limit:
$$\lim_{x \to \infty} \frac{x^3 + 3x^2 + 4x + 5}{x^3}$$
Again, by the same notion, the numerator could just be replaced with $x^3$ to conclude the limit is $1$. In fact, for something of this sort, you could even just multiply the top and bottom by $1/x^3$:
$$\lim_{x \to \infty} \frac{x^3 + 3x^2 + 4x + 5}{x^3} = \lim_{x \to \infty} \frac{1 + 3/x + 4/x^2 + 5/x^3}{1}$$
which, again, makes the limit more obvious. 

If you want to use this (somewhat more rigorous) approach in your original limit, you could utilize the binomial theorem:
$$(a+b)^n = \sum_{k=0}^n \binom n k a^k b^{n-k}$$
In your case, let $a=x^2$ and $b=1$, and replace $n$ with $n+1$. Then pull out the $n+1$ term and we'll see
$$(x^2 + 1)^{n+1} = \sum_{k=0}^{n+1} \binom {n+1} k (x^2)^k 1^{n-k} = x^{2(n+1)} + \sum_{k=0}^{n} \binom {n+1} k x^{2k}$$
Then your limit becomes
$$\lim_{x \to \infty}\frac{x^{2(n+1)}}{(x^2+1)^{n+1}} = \lim_{x \to \infty}\frac{x^{2(n+1)}}{x^{2(n+1)} + \sum_{k=0}^{n} \binom {n+1} k x^{2k}}$$
Now multiply the top and bottom by $1/x^{2(n+1)}$:
$$\lim_{x \to \infty}\frac{x^{2(n+1)}}{x^{2(n+1)} + \sum_{k=0}^{n} \binom {n+1} k  x^{2k}} = \lim_{x \to \infty}\frac{1}{1 + \sum_{k=0}^{n} \binom {n+1} k x^{2k-2(n+1)}}$$
Note that $2k-2(n+1) < 0$ since $k$ is at most $n$. Then $x^{2k-2(n+1)} \to 0$ in the limit, and thus the entire summation goes to $0$.
A: Divide the integral into unit steps as follows:
$$\int_0^{\infty} f(x) dx=\int_0^1 f(x) dx  + \int_1^2 f(x) dx + \int_2^3 f(x) dx + ... $$
If $f(x)$ is monotonically decreasing then we can provide an upper bound for each integral on the RHS (and thus also for the integral on the LHS) using the values of $f(x)$ at the lower limits :
$$\int_0^{\infty} f(x) dx \lt f(0) \Delta x  + f(1) \Delta x + f(2) \Delta x + ...=\sum_{k=0}^{\infty} f(k) $$ where $\Delta x=1$ of course. If we know that the series sum on the RHS converges this can be used to test whether the integral on the LHS converges.
Conversely we can provide a lower bound for each integral on the RHS (and also for the integral on the LHS) using the values of $f(x)$ at the upper limits :
$$\int_0^{\infty} f(x) dx \gt f(1) \Delta x  + f(2) \Delta x + f(3) \Delta x + ...=\sum_{k=1}^{\infty} f(k) $$ If we know that the integral on the LHS converges this can be used as a test for the convergence of the series sum on the RHS.
The last inequality is illustrated in the following graph. 

(Image borrowed from : Paul's Online Notes)
Note that we can deal with other integer limits also :
$$\int_a^b f(x) dx \lt \sum_{k=a}^{b-1} f(k) $$
$$\int_a^b f(x) dx \gt \sum_{k=a+1}^b f(k) $$

To check that your integral converges, first note that $\frac{1}{(x^2+1)^{n+1}}$ is monotonically decreasing provided that $n \gt -1$. Then we can use the 1st inequality above to write that
$$\int_0^{\infty} \frac{1}{(x^2+1)^{n+1}}dx \lt \sum_{k=0}^{\infty} \frac{1}{(k^2+1)^{n+1}} \lt 1+\sum_{k=1}^{\infty} \frac{1}{(k^2)^{n+1}}\lt 1+\sum_{k=1}^{\infty} \frac{1}{k^2} $$ where each denominator is smaller than the corresponding one in the sum to the left, therefore corresponding terms are getting larger and the series sums are getting larger to the right.
If you know that the last series sum is convergent (it is in fact equal to $\frac{\pi^2}{6}$) then you can immediately conclude that the integral on the LHS also converges.
A: You have
$$\frac{x^{2(n+1)}}{(x^2+1)^{n+1}}=\frac{x^{2(n+1)}}{(x^2(1+\frac 1{x^2}))^{n+1}}$$ $$= \frac{x^{2(n+1)}}{x^{2(n+1)}(1+\frac 1{x^2})^{n+1}}=\frac{1}{(1+\frac 1{x^2})^{n+1}}$$ $$\stackrel{x\to \infty}{\longrightarrow}\frac{1}{(1+0)^{n+1}}=1$$
Note, that $\int_{\color{blue}{0}}^{\color{blue}{1}} g(x) \;dx = \int_0^1  \frac 1{x^{2(n+1)}}\; dx = \infty$, but since $\int_{\color{blue}{0}}^{\color{blue}{1}} f(x) \;dx = \int_0^1  \frac 1{(x^2+1)^{n+1}}\; dx < \infty$, the limit comparison test should not be applied to $[\color{blue}{0},+\infty)$ but, for example, to $[\color{blue}{1},+\infty)$:
Now, since 


*

*$f(x), g(x) > 0$ on $[1,+\infty) $ and 

*$\int_{\color{blue}{1}}^{\infty}\frac 1{x^{2(n+1)}}<+\infty$ and

*$\lim_{x\to+\infty}\frac{f(x)}{g(x)} = 1$
you can invoke the limit comparison test and conclude that
$\int_1^{+\infty}  \frac 1{(x^2+1)^{n+1}}\; dx < +\infty$ and, since $\int_0^{1}  \frac 1{(x^2+1)^{n+1}}\; dx <+\infty$, you obtain
$$\int_0^{+\infty}  \frac 1{(x^2+1)^{n+1}}\; dx < +\infty$$
Note, that a direct comparison test would have been much simpler:
$$\int_0^{+\infty}\frac{dx}{(x^2+1)^{n+1}}\leq\int_0^{+\infty}\frac{dx}{x^2+1}$$ $$=\lim_{x\to+\infty}\arctan x = \frac{\pi}2<+\infty$$
