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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.

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    $\begingroup$ Are you sure the answer is $e/2$ instead of $e$? $\endgroup$ – Larry Dec 26 '18 at 22:32
  • $\begingroup$ That's what Wolfram Alpha suggests. I'm not an expert so I don't doubt Wolfram's answers for problems at my level - if it's wrong, sorry for implying that. link to Wolfram $\endgroup$ – Bag of Chips Dec 26 '18 at 22:34
  • $\begingroup$ I see, wolfram alpha indeed gives e/2 $\endgroup$ – Larry Dec 26 '18 at 22:36
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    $\begingroup$ I am getting this... $$\lim_{n \to \infty}n\bigg(\bigg(1+\frac{1}{n}\bigg)^{n+1}-e\bigg)$$ $$=\lim_{n \to \infty}n\bigg(\bigg(1+\frac{1}{n}\bigg)^n\bigg(1+\frac{1}{n}\bigg)-e\bigg)$$ $$=\lim_{n \to \infty}n\bigg(e\bigg(1+\frac{1}{n}\bigg)-e\bigg)$$ $$=\lim_{n \to \infty}n\bigg(e\bigg(1+\frac{1}{n}-1\bigg)\bigg)$$ $$=\lim_{n \to \infty}n(\frac{e}{n})$$ $$=e$$ $\endgroup$ – Rakibul Islam Prince Dec 26 '18 at 23:01
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    $\begingroup$ @RakibulIslamPrince: you cannot replace a part of the expression with its limit in general. $\endgroup$ – Paramanand Singh Dec 27 '18 at 0:31
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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.

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  • $\begingroup$ The question is "why can't I take the partial limit"? $\endgroup$ – Rakibul Islam Prince Dec 27 '18 at 11:01
  • $\begingroup$ @RakibulIslamPrince: think why can't you write $\sqrt{a+b} =\sqrt{a} +\sqrt{b} $. $\endgroup$ – Paramanand Singh Dec 28 '18 at 3:31
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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? :)

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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.

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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 $$\frac{e}{2}-\frac{e}{240}=\frac{119 e}{240}\approx 1.34781$$

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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 $e/2$.

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