Exponential and factorial inequality (1 Viewer)

juantheron · Mar 30, 2023

Proving the result $\displaystyle \frac{n^n}{n!}>2^n, n>6$

cossine · Mar 30, 2023

Try proving n/ (n!)^(1/n) >2

tywebb · Mar 31, 2023

It is also true for n=6.

Induction proof.

$\frac{6^6}{6!}=64.8>64=2^6\therefore \text{true for }n=6$

$\text{If true for }n=k, \frac{k^k}{k!}>2^k$

$\text{and noting that }\frac{(k+1)^{k+1}k!}{k^k(k+1)!}=(1+\frac{1}{k})^k=1+k\cdot\frac{1}{k}+\sum_{r=2}^k{k\choose r}\cdot\frac{1}{k^r}>2$

$\text{then }\frac{(k+1)^{k+1}}{(k+1)!}>2\cdot\frac{k^k}{k!}>2\cdot2^k=2^{k+1}$ .

$\text{Hence it is true for }n=k+1$

Hence by the principal of mathematical induction, it is true for all integers $n\ge6$

juantheron · Mar 31, 2023

cossine said:
Try proving n/ (n!)^(1/n) >2

Thanks cossine.

Using your hint , We use Stirling approximation

$\displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^{n}\sqrt{2\pi n}\approx \bigg(\frac{n}{e}\bigg)^n.$

So we have $\displaystyle \frac{n}{\bigg(n!\bigg)^{\frac{1}{n}}}\approx e>2\Longrightarrow n!<\bigg(\frac{n}{2}\bigg)^n$

juantheron · Mar 31, 2023

tywebb said:
It is also true for n=6.

Induction proof.

$\frac{6^6}{6!}=64.8>64=2^6\therefore \text{true for }n=6$

$\text{If true for }n=k, \frac{k^k}{k!}>2^k$

$\text{and noting that }\frac{(k+1)^{k+1}k!}{k^k(k+1)!}=(1+\frac{1}{k})^k=1+k\cdot\frac{1}{k}+\sum_{r=2}^k{k\choose r}\cdot\frac{1}{k^r}>2$

$\text{then }\frac{(k+1)^{k+1}}{(k+1)!}>2\cdot\frac{k^k}{k!}>2\cdot2^k=2^{k+1}$ .

$\text{Hence it is true for }n=k+1$

Hence by the principal of mathematical induction, it is true for all integers $n\ge6$

Thanks Tywebb.

I have tried like this way

I am assuming $\displaystyle n=2k, k\geq 3, k\in\mathbb{Z}$

So we have $\displaystyle \frac{n!}{\bigg(\frac{n}{2}\bigg)^n}=2\prod^{\frac{n}{2}-1}_{r=1}\frac{\bigg(\frac{n}{2}-r\bigg)\bigg(\frac{n}{2}+r\bigg)}{\frac{n}{2}\cdot \frac{n}{2}}=2\prod^{\frac{n}{2}-1}_{r=1}\bigg[1-\frac{r^2}{\bigg(\frac{n}{2}\bigg)^2}\bigg]<1$

synthesisFR · Mar 31, 2023

juantheron said:
Thanks Tywebb.

I am assuming $\displaystyle n=2k, k\geq 3, k\in\mathbb{Z}$

So we have $\displaystyle \frac{n!}{\bigg(\frac{n}{2}\bigg)^n}=2\prod^{\frac{n}{2}-1}_{r=1}\frac{\bigg(\frac{n}{2}-r\bigg)\bigg(\frac{n}{2}+r\bigg)}{\frac{n}{2}\cdot \frac{n}{2}}=2\prod^{\frac{n}{2}-1}_{r=1}\bigg[1-\frac{r^2}{\bigg(\frac{n}{2}\bigg)^2}\bigg]<1$

U need to stop pulling out the uni stuff for mx2

juantheron said:
Using your hint , We use Stirling approximation

$\displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^{n}\sqrt{2\pi n}\approx \bigg(\frac{n}{e}\bigg)^n.$

So we have $\displaystyle \frac{n}{\bigg(n!\bigg)^{\frac{1}{n}}}\approx e>2\Longrightarrow n!<\bigg(\frac{n}{2}\bigg)^n$

or this

Exponential and factorial inequality (1 Viewer)

juantheron

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cossine

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tywebb

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juantheron

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juantheron

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synthesisFR

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