$(2/3)^{c\log n} < n^{-2}$ when $n>1$ for some $c$. Seems like magic to me that they spotted this (logarithmic manipulation most likely required)

Question

So, in more readable form, the following is true for sufficiently large $c$ and for $n>1$:

$$\left(\frac{2}{3}\right)^{c \log(n)} < n^{-2}$$

This is literally given as a step in the proof I am reading with no explanation or justification. Wolfram Alpha tells me it is correct, but could somebody please break down the mathematics for this step? I am supposed to have been able to spot this step myself.

Answer 1

$$\left(\frac{2}{3}\right)^{c \log(n)} < n^{-2}$$ Taking logarithm on both sides, we get $$\log\left(\frac{2}{3}\right)^{c \log(n)} < \log n^{-2}$$ $$\Rightarrow c \log(n)\cdot \log\left(\frac{2}{3}\right) < -2\log n$$ $$\Rightarrow c\cdot \log\left(\frac{2}{3}\right) < -2$$ Since $n>1$, we have $\log n > 0$, so we can perform the above cancellation without change of inequality. Hence, we have, $$ c > -\frac{2}{\log\left(\frac{2}{3}\right)}$$

Since $\log\left(\frac{2}{3}\right) < 0$, hence the change of the inequality.

Eventually this simplifies to $$c > 11.358$$

Hope this helps.

Answer 2

Taking the logarithm of both sides, this is the same as $$ -ck\log n<-2\log n $$ where $k=\log(3/2)$.

(Note: this assumes that the base of the logarithm is $>1$.)

Answer 3

Hint:

Using logarithms it is equivalent to prove that:

$$ k \log n < -2 \log n $$

with $k=c(\log2-\log 3)$ .

Note that $(\log2-\log 3)<0$ and find $c$.

Answer 4

HINT: your inequality is equivalent to

$$n^2(2/3)^{c\ln n}=n^2n^{c\ln(2/3)}=n^{2+c\ln(2/3)}<1\\\implies (2-c\ln(3/2))\ln n<0$$

for $n>1$ and sufficiently large $c$, where $\ln(3/2)>0$.

Answer 5

Let

$e^a:= 2/3. $

Then $a = \log(2/3).$

Note $a \lt 0.$

$(e^a)^{c\log(n)} =e^{ac\log(n)}= e^{\log(n^{ac})} = n^{ac}.$

Choose $c$ large enough such $ac \lt -2.$