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How can I show, preferably with elementary methods, that $$lim_{nrightarrowinfty} nleft[left(1+frac{1}{n}right)^{n+1} - eright] = frac{e}{2}$$? 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.

With usual Taylor expansions calculate:

$nleft((1+frac 1n)^{n+1}-eright)
\=nleft(exp((n+1)ln(1+frac 1n))-eright)
\=nleft(exp((n+1)(frac 1n-frac 1{2n^2}+o(frac 1{n^2})))-eright)\=nleft(exp(frac 1n-underbrace{frac 1{2n^2}}_{(*)=0}+1-frac 1{2n}+o(frac 1n)))-eright)
\=nleft(exp(1+frac 1{2n}+o(frac 1n)))-eright)
\require{cancel}=nleft(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
$$left(1+frac 1nright)^nleft(1+frac 1nright)-e=(e)(1)-e=0$$

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:

$nleft(exp((n+1)(frac 1n+o(frac 1{n})))-eright)
\=nleft(exp(1+o(1))-eright)
\=n(e+o(1)-e)
\=0+o(n)$

which has not clear limit.

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)logfrac{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? :)

This is a modification of the answer by Sorin Tirc

$$lim_{nrightarrowinfty} nleft[left(1+frac{1}{n}right)^{n+1} - eright] = lim_{nrightarrowinfty} frac{e^{(n+1) ln left(1+frac{1}{n}right)} - e^0}{frac{1}{n}}\= lim_{nrightarrowinfty} frac{e^{(n+1) ln left(1+frac{1}{n}right)} - e^1}{(n+1) ln left(1+frac{1}{n}right)-1}frac{(n+1) ln left(1+frac{1}{n}right)-1}{frac{1}{n}} $$

Now, since
$$lim_{x to 0} frac{e^x-e^1}{x-1}=e$$ by the definition of the derivative you get
$$lim_{nrightarrowinfty} nleft[left(1+frac{1}{n}right)^{n+1} - eright] = lim_{nrightarrowinfty} e frac{(n+1) ln left(1+frac{1}{n}right)-1}{frac{1}{n}}= lim_{nrightarrowinfty} [(n+1) ln left(1+frac{1}{n}right)-1]$$
which can easi;y be calculated with L'Hospital.

Considering$$a_n=nleft(left(1+frac{1}{n}right)^{n+1}-eright) $$ Consider
$$x=left(1+frac{1}{n}right)^{n+1}implies log(x)=(n+1)logleft(1+frac{1}{n}right)$$ So, using Taylor
$$log(x)=(n+1)left(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
n^3}+Oleft(frac{1}{n^4}right)right)=1+frac{1}{2 n}-frac{1}{6 n^2}+Oleft(frac{1}{n^3}right)$$

Continue with Taylor
$$x=e^{log(x)}=e+frac{e}{2 n}-frac{e}{24 n^2}+Oleft(frac{1}{n^3}right)$$ Back to $a_n$
$$a_n=nleft(e+frac{e}{2 n}-frac{e}{24 n^2}+Oleft(frac{1}{n^3}right)-eright)=frac{e}{2}-frac{e}{24 n}+Oleft(frac{1}{n^2}right) $$ 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
$$a_{10}=frac{285311670611}{10000000000}-10 eapprox 1.34835$$ while the approximation would give $$frac{e}{2}-frac{e}{240}=frac{119 e}{240}approx 1.34781$$

Let's put $x=1/n$ so that $xto 0^{+}$ and the expression under limit can be written as $$frac{(1+x)^{1+(1/x)}-e}{x}$$ The numerator can be expressed as $$ecdotdfrac{expleft(dfrac{1+x}{x}log(1+x)-1right)-1}{dfrac{1+x}{x}log(1+x)-1} cdot left(dfrac{1+x}{x}log(1+x)-1right) $$ and the middle fraction tends to $1$ so that the desired limit is equal to the limit of $$ecdotfrac{(1+x)log(1+x)-x}{x^2}$$ which is same as the limit of $$eleft(frac{log(1+x)}{x}+frac{log(1+x)-x}{x^2}right)$$ 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 $e/2$.