Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that 

$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$

I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$
Doing some manipulation we get: 


$$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$


How can I prove this limit?
 A: Note that $\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] = \sum_{k=1}^n \ln\left(1+\frac{k}{n^2}\right)$.
Since for $x\geq 0$, $|\ln(1+x)-x|\leq \frac{x^2}2$, 
$$\left|\sum_{k=1}^n \ln\left(1+\frac{k}{n^2}\right)-\sum_{k=1}^n \frac{k}{n^2} \right|\leq \sum_{k=1}^n \frac{k^2}{2n^4}\leq \frac{1}{2n}$$
But $\displaystyle \sum_{k=1}^n \frac{k}{n^2} = \frac 1n \sum_{k=1}^n \frac{k}{n}$ is a Riemann sum that converges to $\int_0^1 t dt = \frac 12$.
Thus $$\lim_{n\to \infty}\sum_{k=1}^n \ln\left(1+\frac{k}{n^2}\right) = \frac 12$$
A: Via Stirling's formula,
$$
\begin{align}
\frac{(n^2+n)!}{n^2!n^{2n}} &= \frac{\sqrt{2\pi(n^2+n)}\left(\frac{n^2+n}{e}\right)^{n^2+n}}{\sqrt{2\pi n^2}\left(\frac{n^2}{e}\right)^{n^2}n^{2n}}(1+o(1)) \\
&= e^{-n}\left(1+\frac{1}{n}\right)^{n^2+n}(1+o(1))
\end{align}
$$
Taking logarithms,
$$
\begin{align}
\log\frac{(n^2+n)!}{n^2!n^{2n}} &= -n+(n^2+n)\log{\left(1+\frac{1}{n}\right)}+\log(1+o(1)) \\
&=-n+(n^2+n)\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)+o(1) \\
&=\frac{1}{2}+o(1).
\end{align}
$$
A: 
I thought it might be instructive to present an approach that does not rely on calculus, but rather uses the squeeze theorem, a set of inequalities that can be obtained with pre-calculus tools only, and the values of the sums $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$.  To that end we proceed.


TOOL $1$:  Elementary Inequality
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities 
$$\frac{x-1}{x}\le \log(x)\le x-1\tag 1$$

TOOL $2$:  Simple Lower Bound for $\frac{1}{1+x}$
Noting that $1-x^2\le 1$, it is trivial to see that 
$$1-x\le \frac{1}{1+x}\tag 2$$
for $x>-1$.

TOOL $3$:  Values of Sums of Integers and Squared Integers
It is easy to show, using induction for example, or as I showed in THIS ANSWER using elementary analysis, that 
$$\sum_{k=1}^n k=\frac{n(n+1)}{2}\tag 3$$
and 
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}\tag 4$$

Applying the inequalities in $(1)$ with $x=\frac{k}{n^2}$, we find that 
$$\begin{align}
\frac{\frac{k}{n^2}}{1+\frac{k}{n^2}}\le \log\left(1+\frac k{n^2}\right)\le \frac{k}{n^2}\tag 5
\end{align}$$

Next, applying the inequality $(2)$ to the left side of $(5)$ with $x=\frac{k}{n^2}$ reveals
$$\begin{align}
\frac{k}{n^2}-\frac{k^2}{n^4}\le \log\left(1+\frac k{n^2}\right)\le \frac{k}{n^2}\tag 6
\end{align}$$

Summing the terms in $(6)$ from $k=1$ to $k=n$ yields
$$\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^2}{n^4}\right)\le \sum_{k=1}^n \log\left(1+\frac k{n^2}\right)\le \sum_{k=1}^n \frac{k}{n^2}\tag 7$$
whereupon applying $(3)$ and $(4)$ to $(7)$ we find that 
$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^4}\le \sum_{k=1}^n \log\left(1+\frac k{n^2}\right)\le \frac{n(n+1)}{2n^2}\tag 8$$

Finally, applying the squeeze theorem to $(8)$, we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\log\left(1+\frac k{n^2}\right)=\frac12}$$

A: From $x-{1\over2}x^2\lt\log(1+x)\lt x$ for $0\lt x\lt1$ and 
$${k\over n^2}-{k\over n(n+1)}={k\over n^2(n+1)}$$
we find
$$0\lt\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\lt{k\over n^2(n+1)}$$
for $1\le k\le n$. Thus
$$0\lt\sum_{k=1}^n\left(\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\right)=\sum_{k=1}^n\log\left(1+{k\over n^2}\right)-{1\over2}\lt\sum_{k=1}^n{k\over n^2(n+1)}={1\over2n}\to0$$
Remark: The $0$ lower bound comes from
$${k\over n^2(n+1)}-{k^2\over2n^4}\gt{k\over n^2(n+1)}-{k\over2n^3}$$
A: HINT
$$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)=\frac{1}{n^2}\log\Big(1+\frac 1{n^2}\Big)^{n^2}+ \frac{2}{n^2}\log\Big(1+\frac 2{n^2}\Big)^\frac{n^2}{2}+...+ \frac{n}{n^2}\log\Big(1+\frac n{n^2}\Big)^\frac{n^2}{n}=\frac{1}{n^2}\sum_1^n k\log\left(1+\frac{k}{n^2}\right)^\frac{n^2}{k}\sim \frac{1}{n^2}\sum_1^n k=\frac{1}{n^2}\frac{n(n+1)}{2} \to \frac  12$$
