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?
Thanks in advance
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1bounds for what sort of $n$? – user10354138 Sep 30 '18 at 06:51
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when n is positive – manifold Sep 30 '18 at 06:53
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Huh? For example, which $\log(1/2)$-power of $-1$? (when $n=\frac12>0$) – user10354138 Sep 30 '18 at 07:16
2 Answers
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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 $$ \left( \frac{1}{n} \right)^{\frac{1}{n-1}} \leq f(n) \leq \left( \frac{1}{n} \right)^{\frac{1}{n}}. $$
Hugo
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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 $1$: $$f(n)=\mathrm e^{\log n\,\log\bigl(1-\tfrac1n\bigr)}=\mathrm e^{\log n\bigl(-\tfrac1n+o\bigl(\tfrac1n\bigr)\bigr)}=\mathrm e^{-\tfrac{\log n}n+o\bigl(\tfrac{\log n}n\bigr)}\to\mathrm e^0.$$
Bernard
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