# lower bound and upper bound of expression ($1-\frac{1}{n})^{\log n}$

Let's say that $$f(n)=\bigl(\frac{n-1}{n}\bigr)^{\log n}$$ ( I know bounds of $$\bigl(\frac{n-1}{n}\bigr)^{n}$$ ).

Is there any way I can get good upper bound of $$f(n)$$ when n is positive?

• bounds for what sort of $n$? – user10354138 Sep 30 '18 at 6:51
• Huh? For example, which $\log(1/2)$-power of $-1$? (when $n=\frac12>0$) – user10354138 Sep 30 '18 at 7:16
Recall that $$\left( 1 - \frac{1}{n} \right)^n \leq \frac{1}{e} \leq \left( 1 - \frac{1}{n} \right)^{n-1}.$$
Then $$\log f(n) = \log n \cdot \log \left( 1 - \frac{1}{n} \right) \leq -\frac{\log n}{n}$$ and $$\log f(n) = \log n \cdot \log \left( 1 - \frac{1}{n} \right) \geq -\frac{\log n}{n-1},$$ hence $$\left( \frac{1}{n} \right)^{\frac{1}{n-1}} \leq f(n) \leq \left( \frac{1}{n} \right)^{\frac{1}{n}}.$$
For $$n>1$$, $$\;0<1-\dfrac1n<1\;$$ and $$\; \log n>0$$, so $$\Bigl(1-\frac1n\Bigr)^{\log n}<\Bigl(1-\frac1n\Bigr)^0=1.$$ Furthermore, the limit of $$f(n)$$ isqual to $$1$$: $$f(n)=\mathrm e^{\log n\,\log\bigl(1-\tfrac1n\bigr)}=\mathrm e^{\log n\bigl(-\tfrac1n+o\bigl(\tfrac1n\bigr)\bigr)}=\mathrm e^{-\tfrac{\log n}n+o\bigl(\tfrac{\log n}n\bigr)}\to\mathrm e^0.$$