How to prove limit of integral is $\frac{1}{n2^n}$? I was wondering how to prove that $$\lim_{n\to \infty}\int_{1}^{n}\frac{1}{(x^{2}+1)^{n}}dx\sim \frac{1}{n\cdot 2^{n}}?.$$
This appears to be asymptotic to $\frac{1}{n2^{n}}$, but how to prove it?.
I checked with larger and larger values of n, and it does get closer and closer to 
$$\frac{1}{n2^{n}}.$$  
i.e 
$$\int_{1}^{10}\frac{1}{(x^{2}+1)^{10}}dx\approx .00009843725636$$
and 
$$\frac{1}{10\cdot 2^{10}}\approx .00009765625.$$
The larger $n$, the closer they get. 
I tried using parts to no avail. I also thought $$\sum_{k=0}^{\infty}\binom{-n}{k}x^{2k}=\frac{1}{(1+x^{2})^{n}}$$ may be useful in some manner.
Does anyone have a good idea as to how to prove this?
Thanks
 A: Divide $\int_{1}^{n}\frac{2^n}{(x^{2}+1)^{n}}dx$ between $1$ and $1+n^{-2/3}$ and between $1+n^{-2/3}$ and $n$.
The second part is less than $n \left( \frac{2}{(1+n^{-2/3})^2+1}\right)^n$ which is equivalent to $n \exp (n(-n^{-2/3}+o(n^{-2/3}))$ which can be neglected considering the equivalent we are going to find for the other part.
For the first part, we use $\frac{2}{(1+u)^2+1}=1-u+O(u^2)$ as $u \rightarrow 0$ to evaluate $\int_0^{n^{-2/3}}\left(\frac{2}{(1+u)^2+1}\right)^n du$.
This way we get that $e^{-nu-Cnu^2} \leq \left(\frac{2}{(1+u)^2+1}\right)^n \leq e^{-nu+Cnu^2}$ for some constant $C \gt 0$ and for all $u \in [0,1]$.
So the first part falls between $e^{-Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$ and $e^{Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$, so is equivalent to $\int_0^{n^{-2/3}} e^{-nu} du = \frac{1}{n}(1-e^{-n^{1/3}}) \sim 1/n$.
$n^{-2/3}$ could be replaced by any $\epsilon_n$ such that $n \epsilon_n \rightarrow + \infty$ and $n \epsilon_n^2 \rightarrow 0$
A: Since $\int_2^n {\frac{1}{{(x^2  + 1)^n }}\,{\rm d}x}  \le \frac{{n - 2}}{{5^n }}$, it suffices to show that
$$
\frac{1}{{(n + 1)2^n }} \le \int_1^2 {\frac{1}{{(x^2  + 1)^n}}\,{\rm d}x}  \le \frac{1}{{(n - 1)2^n }}.
$$
This follows straight from
$$
\frac{{2 - x}}{2} \le \frac{1}{{x^2  + 1}} \le \frac{1}{{2x}}.
$$
A: Motivated by the simplicity of integrating from $0$, rather than $1$, substituting $x = 1+u$ gives
$$2^n\int_1^n{\frac{dx}{(1+x^2)^n}} = \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}}.$$
We seek estimates--an upper bound and a lower bound for the integrand--that will be simple to integrate.  Dropping the "complicated" $u^2/2$ term gives one obviously integrable estimate; continuing the pattern $1+u+u^2/2 + \cdots + u^k/k! + \cdots = e^u$ leads to another.  The resulting inequalities $1 + u < 1 + u + u^2/2 < e^u$ yield (for $n \ge 1$)
$$\eqalign{
\frac{1}{n}\left(1 - e^{-n(n-1)}\right) &= \int_0^{n-1}{e^{-n u}du} \cr
&\lt \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}} \cr
&\lt \int_0^{n-1}{\frac{du}{(1+u)^n}} = \frac{1}{n-1}\left(1 - \frac{1}{n^{n-1}}\right) \cr
&\lt \frac{1}{n-1}.
}$$
Both estimates are asymptotically $1/n$, QED.
A: This one should be a comment.
As PEV said you could use contour integration to show that $\int_{0}^{1} \frac{1}{(1+x^2)^n} dx$ is of the same order as $\int_{1}^{n} \frac{1}{(1+x^2)^n} dx$
To evaluate it directly you could try the following.
Plug in $x = \tan(\theta)$.
The integral becomes $\displaystyle \int_{\pi/4}^{\tan^{-1}n} \frac{\sec^2 (\theta)}{\sec^{2n} (\theta)} d \theta = \int_{\pi/4}^{\tan^{-1}n} \cos^{2n-2}(\theta) d \theta \approx \int_{\pi/4}^{\pi/2} \cos^{2n-2}(\theta) d \theta = \frac{\text{(Hypergeometric function)}}{2^{n-1}(2n-1)}$.
The approximation tends to better and better asymptotically since $\cos(\theta)$ is bounded.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
I_{n} \equiv \int_{1}^{n}{\dd x \over \pars{x^{2} + 1}^{n}}&=
{n \over \pars{n^{2} + 1}^{n}}
- {1 \over 2^{n}} + 2n\int_{1}^{n}{x^{2} \over \pars{x^{2} + 1}^{n + 1}}\,\dd x
\\[3mm]&=
{n \over \pars{n^{2} + 1}^{n}}
- {1 \over 2^{n}} + 2nI_{n} - 2n\int_{1}^{n}{\dd x \over \pars{x^{2} + 1}^{n + 1}}
\end{align}

\begin{align}
I_{n} &=
{n \over \pars{1 - 2n}\pars{n^{2} + 1}^{n}}
+ \color{#00f}{{1 \over \pars{2n - 1}2^{n}}}  + {2n \over 2n - 1}\,\bracks{%
I_{n + 1} - \int_{n}^{n + 1}{\dd x \over \pars{x^{2} + 1}^{n + 1}}
}
\end{align}
  The $\color{#00f}{\large blue}$ one is 'the leading term':
  $$
{1 \over \pars{2n - 1}2^{n}} \sim {1 \over n2^{n + 1}} \sim 
\color{#00f}{\large{1 \over n\,2^{n}}}
$$

