Determine $\lim_{n\to \infty} n\left(1+(n+1)\ln \frac{n}{n+1}\right)$ Determine $$\lim_{n\to \infty} n\left(1+(n+1)\ln \frac{n}{n+1}\right)$$
I noticed the indeterminate case $\infty \cdot 0$ and I tried to get them all under the $\ln$, but it got more complicated and I reached another indeterminate form. The same happened when I tried to use Stolz-Cesaro.
EDIT: is there an elementary solution, without l'Hospital or Taylor series?
 A: Let $\frac {n}{n+1}=e^x.$ As $n\to \infty,\;\; x\to 0$ through negative values.The expression is $$e^x\cdot \frac {1+x-e^x}{(1-e^x)^2}.$$ Applying l'Hopital's Rule to $\frac {1+x-e^x}{(1-e^x)^2}$ we get a limit of $\frac {-1}{2}$ as $x\to 0.$ And the far left term $e^x$ in the expression goes to $1$ as $x\to 0.$
A: $$\ln\dfrac{n}{n+1}=\ln(1-\dfrac{1}{n+1})\sim-\dfrac{1}{n+1}-\dfrac{1}{2(n+1)^2}-\mathcal{O}(\dfrac{1}{n^3})$$
then
$$1+(n+1)\ln\dfrac{n}{n+1}\sim-\dfrac{1}{2(n+1)}-\mathcal{O}(\dfrac{1}{n^2})$$
and
$$n(1+(n+1)\ln\dfrac{n}{n+1})\sim-\dfrac{n}{2(n+1)}-\mathcal{O}(\dfrac1n)\to-\dfrac12$$
as $n\to\infty$.
A: Compute the limit at infinity of
$$
f(x)=(x-1)\left(1+x\ln\frac{x-1}{x}\right)
$$
which, with $t=1/x$, is the same as
$$
\lim_{t\to0^+}\frac{1-t}{t}\left(1+\frac{\ln(1-t)}{t}\right)=
\lim_{t\to0^+}\frac{t+\ln(1-t)}{t^2}=
\lim_{t\to0^+}\frac{t-t-t^2/2+o(t^2)}{t^2}=-\frac{1}{2}
$$
The factor $1-t$ can be disregarded, as it has limit $1$ and the remaining factor has a limit.
The original limit is
$$
\lim_{n\to\infty}f(n+1)
$$
A: Upgrading my comment to an answer since I think I actually believe in it:
 A Taylor Series expansion near $x=0$ gives
$$\log(1-x)=-x-\frac{1}{2}x^{2}+o(x^{2})$$
as $x \to 0$. Hence, taking $x=\frac{1}{n+1}$, 
$$1+(n+1)\log\left(1-\frac{1}{n+1}\right)=1-\frac{n+1}{n+1}-\frac{n+1}{2(n+1)^{2}}+o(n^{-1})$$
and, multiplying by $n$,
$$n\left(1+(n+1)\log\left(1-\frac{1}{n+1}\right)\right)=-\frac{n}{2(n+1)}+o(1)$$
which safely tends to $-1/2$.
A: My recommendation is to always use asymptotic expansion. But if you want purely elementary means, these inequalities are helpful and can be proven easily:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

  
*
  
*$\ln(1+x) \ge \lfrac{2x}{2+x}$ for every real $x \ge 0$.
  
*$\ln(1+x) \le x-\lfrac12x^2+\lfrac13x^3$ for every real $x \ge 0$.
  
*$\ln(1+x) \le \lfrac{2x}{2+x}$ for every real $x \in (-1,0]$.
  
*$\ln(1+x) \ge \lfrac{x}{1+x}+\lfrac12(\lfrac{x}{1+x})^2+\lfrac13(\lfrac{x}{1+x})^3$ for every real $x \in (-1,0]$.

The key is the following fact:

Given any functions $f,g$ on $[0,c]$ where $c \ge 0$, if $f(0) = g(0)$ and $f' \le g'$ then $f \le g$.
Given any functions $f,g$ on $[c,0]$ where $c \le 0$, if $f(0) = g(0)$ and $f' \le g'$ then $f \ge g$.

So to prove (1) and (3) it suffices to prove $\lfrac1{1+x} \ge \lfrac{4}{(2+x)^2}$ for real $x > -1$, which is trivial.
And to prove (2) it suffices to prove $\lfrac1{1+x} \le 1-x+x^2$ for real $x > 0$, which is also trivial.
And (4) immediately falls out from the identity $\ln(1+x) = -\ln(1+\lfrac{-x}{1+x})$ and (2).
Incidentally, the proof for (2) shows that you can easily prove the Taylor series for $\ln(1+x)$ for $x > 0$ by elementary means, since it is an alternating series that converges iff $x < 1$.

Now take any natural $n$.
If $x = -\lfrac1{n+1}$ then $\lfrac{x}{1+x} = -\lfrac1n$ and $\lfrac{2x}{2+x} = -\lfrac{2}{2n+1}$. Thus (3) and (4) immediately give:
  $-\lfrac1{n}+\lfrac1{2n^2}-\lfrac1{3n^3} \le \ln(\lfrac{n}{n+1}) \le -\lfrac{2}{2n+1}$.
Therefore we get:
  $n+(n+1)·(-1+\lfrac1{2n}-\lfrac1{3n^2}) \le n·(1+(n+1)·\ln(\lfrac{n}{n+1})) \le \lfrac{-n}{2n+1}$.
And squeeze theorem does the rest.
A: Suppose that $n\le x\le n+1$. Let $u=x-\left(n+\frac12\right)$. Then $-\frac12\le u\le\frac12$ and
$$
\begin{align}
x(2n+1-x)
&=\left(\left(n+\tfrac12\right)+u\right)\left(\left(n+\tfrac12\right)-u\right)\\
&=\left(n+\tfrac12\right)^2-u^2\\
&\in\left[\left(n+\tfrac12\right)^2-\tfrac14,\left(n+\tfrac12\right)^2\right]\tag1
\end{align}
$$
Therefore,
$$
\begin{align}
\log\left(\frac{n+1}n\right)
&=\int_n^{n+1}\frac1x\,\mathrm{d}x\tag{2a}\\
&=\frac12\int_n^{n+1}\left(\frac1x+\frac1{2n+1-x}\right)\mathrm{d}x\tag{2b}\\
&=\int_n^{n+1}\frac{n+\frac12}{x(2n+1-x)}\,\mathrm{d}x\tag{2c}\\
&\in\left[\frac1{n+\frac12},\frac1{n+\frac12-\frac1{4n+2}}\right]\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: a characterization of $\log$
$\text{(2b)}$: average $\text{(2a)}$ with the substitution $x\mapsto2n+1-x$
$\text{(2c)}$: simplify
$\text{(2d)}$: apply $(1)$
Applying $(2)$, we get
$$
\begin{align}
n\left(1+(n+1)\log\left(\frac{n}{n+1}\right)\right)
&\in\left[n\left(1-\frac{n+1}{n+\frac12-\frac1{4n+2}}\right),n\left(1-\frac{n+1}{n+\frac12}\right)\right]\\
&=\left[n\left(\frac{-\frac12-\frac1{4n+2}}{n+\frac12-\frac1{4n+2}}\right),n\left(\frac{-\frac12}{n+\frac12}\right)\right]\\
&=\left[\color{#C00}{\frac{n}{n+\frac12-\frac1{4n+2}}}\color{#090}{\left(-\frac12-\frac1{4n+2}\right)},\color{#C00}{\frac{n}{n+\frac12}}\color{#090}{\left(-\frac12\right)}\right]\tag3
\end{align}
$$
By the Squeeze Theorem, we get
$$
\begin{align}
\lim_{n\to\infty}n\left(1+(n+1)\log\left(\frac{n}{n+1}\right)\right)
&=\color{#C00}{1}\cdot\color{#090}{\left(-\frac12\right)}\\
&=-\frac12\tag4
\end{align}
$$
