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Find $\displaystyle\lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}$

We have: $$ \lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}= \lim_{n\to\infty}\dfrac{e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)}}{e^n}= \lim_{n\to\infty} e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)-n} $$

Then I have no idea. Can anyone help me please?

egreg
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liverpool29
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    This has already been answered here: http://math.stackexchange.com/questions/360510/what-is-lim-n-to-infty-frac1en-bigl1-frac1n-bigrn2?rq=1 – DoppeDee Oct 13 '16 at 11:06

2 Answers2

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The idea of reducing everything to a single exponential is good. Now you have to evaluate $$ \lim_{n\to\infty}\Bigl(n^2\ln\Bigl(1+\frac{1}{n}\Bigr)-n\Bigr) $$ You can try computing instead (with $1/x=t$ after the equality) $$ \lim_{x\to\infty}\Bigl(x^2\ln\Bigl(1+\frac{1}{x}\Bigr)-x\Bigr) = \lim_{t\to0^+}\frac{\ln(1+t)-t}{t^2}= \lim_{t\to0^+}\frac{t-t^2/2+o(t^2)-t}{t^2}=-\frac{1}{2} $$ Instead of a Taylor expansion you can use l'Hôpital: $$ \lim_{t\to0^+}\frac{\ln(1+t)-t}{t^2}= \lim_{t\to0^+}\frac{\dfrac{1}{1+t}-1}{2t} $$

egreg
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$$n^2\log\left(1+\frac{1}{n}\right)-n = \int_{n}^{n+1}\frac{n(n-x)}{x}\,dx=\int_{0}^{1}\frac{-z}{1+\frac{z}{n}}\,dz$$ hence by the dominated convergence theorem $$ \lim_{n\to +\infty}\left[n^2\log\left(1+\frac{1}{n}\right)-n\right]=\int_{0}^{1}-z\,dz = \color{red}{-\frac{1}{2}}.$$

Jack D'Aurizio
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