Evaluating $\lim_{n\rightarrow\infty} n[(1+\frac{1}{n})^{n+1} - e]$ How can I show, preferably with elementary methods, that $$\lim_{n\rightarrow\infty} n\left[\left(1+\frac{1}{n}\right)^{n+1} - e\right] = \frac{e}{2}$$? All that would suffice for me is to show that the limit exists and is not negative. I've tried toying with binomial expansion but it didn't amount to anything unfortunately.
 A: With usual Taylor expansions calculate:

$n\left((1+\frac 1n)^{n+1}-e\right)
\\=n\left(\exp((n+1)\ln(1+\frac 1n))-e\right)
\\=n\left(\exp((n+1)(\frac 1n-\frac 1{2n^2}+o(\frac 1{n^2})))-e\right)\\=n\left(\exp(\frac 1n-\underbrace{\frac 1{2n^2}}_{(*)=0}+1-\frac 1{2n}+o(\frac 1n)))-e\right)
\\=n\left(\exp(1+\frac 1{2n}+o(\frac 1n)))-e\right)
\\\require{cancel}=n\left(\cancel{e}+\frac e{2n}+o(\frac 1n)-\cancel{e}\right)
\\=\frac e{2}+o(1)\to \frac e2
$
(*) this term is too small for the resulting $o(\frac 1n)$ thus it is simply ignored in this expansion.

About the comment of Rakibul Islam Prince:
What is wrong is that you take the limit of $(1+\frac 1n)^n$ inside the calculation. Have you noticed the limit operator is exterior. 
You cannot take partial limits as you wish. 
In fact doing this is equivalent to ignoring the term $-\frac 1{2n}$ coming from order $2$ in log expansion in my calculation.
In essence if you take the limit inside this would give 
$$\left(1+\frac 1n\right)^n\left(1+\frac 1n\right)-e=(e)(1)-e=0$$
Notice the second $1+\frac 1n$ actually cannot stand as is.
Indeed, the correct calculation of expanding the logarithm to first order only would give:
$n\left(\exp((n+1)(\frac 1n+o(\frac 1{n})))-e\right)
\\=n\left(\exp(1+o(1))-e\right)
\\=n(e+o(1)-e)
\\=0+o(n)$ 
which has not clear limit.
A: Another (simple) solution involves applying L’Hopital after using $x=\frac{1}{n}$ to deduce that your limit is equivalent to $\frac{((e^{(1+x)log\frac{1}{x}}-e)}{x} \to \frac{e}{2} $which after differentiating becomes equivalent to $\frac{\frac{x}{x+1}-log(1+x)}{x^2} \to \frac{1}{2}$, for which you apply L’Hopital again and it becomes easy to evaluate, But I assume L’Hopital isn’t really too elementary either? :)
A: This is a modification of the answer by Sorin Tirc
$$\lim_{n\rightarrow\infty} n\left[\left(1+\frac{1}{n}\right)^{n+1} - e\right] = \lim_{n\rightarrow\infty} \frac{e^{(n+1) \ln \left(1+\frac{1}{n}\right)} - e^0}{\frac{1}{n}}\\= \lim_{n\rightarrow\infty} \frac{e^{(n+1) \ln \left(1+\frac{1}{n}\right)} - e^1}{(n+1) \ln \left(1+\frac{1}{n}\right)-1}\frac{(n+1) \ln \left(1+\frac{1}{n}\right)-1}{\frac{1}{n}} $$
Now, since 
$$\lim_{x \to 0} \frac{e^x-e^1}{x-1}=e$$ by the definition of the derivative you get
$$\lim_{n\rightarrow\infty} n\left[\left(1+\frac{1}{n}\right)^{n+1} - e\right] = \lim_{n\rightarrow\infty} e \frac{(n+1) \ln \left(1+\frac{1}{n}\right)-1}{\frac{1}{n}}= \lim_{n\rightarrow\infty} [(n+1) \ln \left(1+\frac{1}{n}\right)-1]$$
which can easi;y be calculated with L'Hospital.
A: Considering$$a_n=n\left(\left(1+\frac{1}{n}\right)^{n+1}-e\right) $$ Consider
$$x=\left(1+\frac{1}{n}\right)^{n+1}\implies \log(x)=(n+1)\log\left(1+\frac{1}{n}\right)$$ So, using Taylor
$$\log(x)=(n+1)\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3
   n^3}+O\left(\frac{1}{n^4}\right)\right)=1+\frac{1}{2 n}-\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right)$$
Continue with Taylor
$$x=e^{\log(x)}=e+\frac{e}{2 n}-\frac{e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ Back to $a_n$
$$a_n=n\left(e+\frac{e}{2 n}-\frac{e}{24 n^2}+O\left(\frac{1}{n^3}\right)-e\right)=\frac{e}{2}-\frac{e}{24 n}+O\left(\frac{1}{n^2}\right) $$ which shows the limit and also how it is approached.
Try with $n=10$ (which is far away from infinity) and use your pocket calculator
$$a_{10}=\frac{285311670611}{10000000000}-10 e\approx 1.34835$$ while the approximation would give $$\frac{e}{2}-\frac{e}{240}=\frac{119 e}{240}\approx 1.34781$$
A: Let's put $x=1/n$ so that $x\to 0^{+}$ and the expression under limit can be written as $$\frac{(1+x)^{1+(1/x)}-e}{x}$$ The numerator can be expressed as $$e\cdot\dfrac{\exp\left(\dfrac{1+x}{x}\log(1+x)-1\right)-1}{\dfrac{1+x}{x}\log(1+x)-1} \cdot \left(\dfrac{1+x}{x}\log(1+x)-1\right) $$ and the middle fraction tends to $1$ so that the desired limit is equal to the limit of $$e\cdot\frac{(1+x)\log(1+x)-x}{x^2}$$ which is same as the limit of $$e\left(\frac{\log(1+x)}{x}+\frac{\log(1+x)-x}{x^2}\right)$$ The first fraction in parenthesis tends to $1$ while the second one tends to $-1/2$ (via an easy application of L'Hospital's Rule or Taylor series) so that the desired limit is $e/2$.
