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This is exercise from Guidorizzi's book Cálculo (In Portuguese).

Show that $\lim_{n\rightarrow \infty}n \big[1-\frac{(n+1)^{n}}{en^{n}}\big]=\frac{1}{2}.$

All I managed to do is rewrite the equation as $n \big[1-\frac{(1+\frac{1}{n})^{n}}{e}\big]$.

What to do from now on?

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Interesting. I've never seen this one before. – Michael Hardy Feb 10 '12 at 22:37
up vote 2 down vote accepted

$$\log\left(1 + \frac{1}{n}\right)^n = n \log \left(1+\frac{1}{n}\right) = n\left(\frac{1}{n}-\frac{1}{2n^2} + O\left(\frac{1}{n^3}\right)\right)$$ Thus $$\left(1 + \frac{1}{n}\right)^n =e^{1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)} $$ Substituting, we need to find $$\lim_{n \to \infty} n\left(1 - \frac{e^{1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)}}{e}\right) = n\left(1-e^{\frac{-1}{2n} + O\left(\frac{1}{n^2}\right)}\right)$$ Now using the expansion of $e^x$, you get $$n\left(1-e^{\frac{-1}{2n} + O\left(\frac{1}{n^2}\right)}\right) = n\left(\frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right)$$ which tends to $\frac{1}{2}$ and $n \to \infty$.

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Hint: $(1+\frac1n)^n=e^{n\log(1+\frac1n)}=e^{1-\frac1{2n}+O(\frac1{n^2})}$.

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