My incorrect approach solving this limit. What am I missing? I'm doing this limit:
$$-1/2\lim_{n\to\infty}\,{\frac {n\left( 2\,{{\rm e}^{2}} \left( {\frac {n+1}{n}}
 \right) ^{-2\,n}n-1-2\,n \right)}{n+1} }
$$
I arrived to this from a bigger expression, and arrived at this point, according to maple, the limit is correctly done. The result (again, according to maple) is $-1/2$, so $$\lim_{n\to\infty}\,{\frac {n\left( 2\,{{\rm e}^{2}} \left( {\frac {n+1}{n}}
 \right) ^{-2\,n}n-1-2\,n \right)}{n+1} }=1.$$
But I've done it several times, and I have obtained $-1$ every time. Here's one of my tryings:

$$\lim_{n\to\infty}\,{\frac {n\left( 2\,{{\rm e}^{2}} \left( {\frac {n+1}{n}}
 \right) ^{-2\,n}n-1-2\,n \right)}{n+1} }$$
We have that
$$\,{{\rm e}^{2}} \left( {\frac {n+1}{n}}
 \right) ^{-2\,n}\to e^2e^{-2}=1$$
So $$\lim_{n\to\infty}\,{\frac {n\left( 2\,{{\rm e}^{2}} \left( {\frac {n+1}{n}}
 \right) ^{-2\,n}n-1-2\,n \right)}{n+1} }=\lim_{n\to\infty}\,{\frac {n\left( 2n-1-2\,n \right)}{n+1} }=\lim_{n\to\infty}\,{\frac {\left( -n \right)}{n+1} }=-1$$

It's obvious that I'm doing a mistake in one of those steps, but I'm not being able to find what it is.
Edit: this is from my analysis course, so I can't use Taylor to solve this limit.
 A: Before the edit
I think that you went too fast.
Consider $$A=\left( {\frac {n+1}{n}}
 \right) ^{-2\,n}$$ Then $$\log(A)=-2n\log(1+\frac 1 n)=-2n\Big(\frac{1}{n}-\frac{1}{2 n^2}+\cdots \Big)=-2+\frac{1}{n}+\cdots$$ So $$A=e^{-2}e^{1/n}=e^{-2}\Big(1+\frac{1}{n}+\cdots  \Big)$$
I am sure that you can take from here.
A: One can show this from first principles. But we need to prove: 
$$\lim_{n \to \infty} n \left(e - \left(1+\frac{1}{n}\right)^n \right) = \frac{1}{2}e$$
Consider the terms in the inner expression. First term:
$$e^1 = 1 + 1/1! + 1/2! + 1/3! + \dots$$
Second term:
\begin{align*}
\left(1+\frac{1}{n}\right)^n &= 1 + n \cdot \frac{1}{n} + \frac{n(n-1)}{2!} \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!} \frac{1}{n^3} + \cdots \\
&= 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right) + \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) + \cdots
\end{align*}
I'm skipping some steps, but with careful subtraction, one can show that:
\begin{align*}
 & n \left(e - \left(1+\frac{1}{n}\right)^n \right) \\
&= \frac{1}{2} (1 + 1/1! + + 1/2! + 1/3! + \dots) + O\left(\frac{1}{n}\right) \\
&\to \frac{1}{2}e
\end{align*}
Now we are all set to prove the actual result. Write:
$$\frac {n\left( \frac{2e^2 n}{\left(1+\frac{1}{n}\right)^{2n}} - 1 - 2n \right)}{n+1} = 
\frac {\left( \left( \frac{2e^2 n}{\left(1+\frac{1}{n}\right)^{2n}} - 2n \right) - 1 \right)}{1+1/n} $$
what we're really concerned with is the inner parentheses of the numerator, since the other part and the denominator have finite limit. Write the inner parentheses as:
$$n \left( \frac{2e^2 - 2 \left(1+\frac{1}{n}\right)^{2n}}{\left(1+\frac{1}{n}\right)^{2n}}\right)$$
again the denominator has a finite limit $e^2$. So, focus only on the numerator:
\begin{align*}
&n \left( 2e^2 - 2 \left(1+\frac{1}{n}\right)^{2n}\right) \\
&= 2 \left( e + \left(1+\frac{1}{n}\right)^{n}\right) \times n\left( e - \left(1+\frac{1}{n}\right)^{n}\right)
\end{align*}
the first part goes to $4e$ and the second part goes to $e/2$ as we proved before. Hence the above expression goes to $2e^2$ and plugging it back we get $1$, QED.
