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I met this problem:

prove:

$\displaystyle \frac{n^n}{3n!}<\frac{e^n}{2}-\sum_{k=0}^{n-1}\frac{n^k}{k!}<\frac{n^n}{2n!}$

I tried expand $e^n$ at $x=0$ then:

$\displaystyle e^n=\sum_{k=0}^{n}\frac{n^k}{k!}+\frac1{n!}\int_0^ne^t(n-t)^ndt$

but I have no idea next...

And another way of thinking:both side divided $e^n$ the LHS became $\displaystyle \frac{n^n}{3n!e^n}$ then can we use the stirling's fomula to find sth.?

Could someone help me? Thanks!

Martin Sleziak
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smallsmallice
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2 Answers2

2

A re-worked answer. We have: $$ e^n = \sum_{k=0}^{n}\frac{n^k}{k!}+\frac{n^{n+1}}{n!}\int_{0}^{1}\left(e^t(1-t)\right)^n\,dt\tag{1}$$ but we also have: $$\forall x\in[0,1],\qquad (1-x^3)e^{-x^2/2} \leq e^{x}(1-x) \leq e^{-x^2/2}\tag{2}$$ hence: $$ \int_{0}^{1}\left(e^t(1-t)\right)^n\,dt \leq \int_{0}^{+\infty}e^{-nx^2/2}\,dx = \sqrt{\frac{\pi}{2n}}\tag{3}$$ and for every $n\geq 11$: $$\begin{align*} \int_{0}^{1}\left(e^t(1-t)\right)^n\,dt \geq \int_{0}^{1}(1-t^3)^n e^{-nt^2/2}\,dt &\geq \int_{0}^{1}(1-nt^3) e^{-nt^2/2}\,dt\\&\geq \int_{0}^{1}e^{-nt^2/2}\,dt-\frac{2}{n}\\&\geq \sqrt{\frac{\pi}{2n}}-\frac{4}{n}.\end{align*}\tag{4}$$ We may go through these lines, improving a bit $(3)$ and $(4)$, or use some probabilistic argument, since we are essentially estimating the probability that a Poisson distributed random variable exceeds its mean value.

Jack D'Aurizio
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  • Could you give me more hints....I think you use $nt=x$ to estimate the remainder.But $e^n$ should multiply by 1/2...and the part of power series still in the fomula? for $e^{-nx^2/2}$ why we calculate $\int_0^\infty$ but not $\int_0^1$ ? – smallsmallice Oct 03 '15 at 00:46
  • @Jack - this doesn't seem to be enough. Rearranging everything I get $$-1/6\leq\big(\frac{1}{K(n)}n\int_0^1(e^x(1-x))^n dx-\frac 1 2\big)K(n)\leq 0$$ where $K(n)=\frac{n!e^n}{n^n}$. Integral behaves as $\sqrt\frac{\pi} {2n}$, $K(n)\sim \sqrt{2\pi n}$, but it's not enough. – A.S. Oct 03 '15 at 09:55
  • @A.S. Laplace's method has to work, it is the standard tool in proving the CLT. If the difference between $\frac{1}{2}$ and $\frac{n}{K(n)}\int_{0}^{1}(e^x(1-x))^n,dx $ is $O\left(\frac{1}{\sqrt{n}}\right)$, we proved something stronger than the original claim. – Jack D'Aurizio Oct 03 '15 at 10:00
  • But it gets multiplied by $K(n)$ which is $O(\sqrt n)$ – A.S. Oct 03 '15 at 10:01
  • @A.S. Eh? $$ \frac{1}{6K(n)}\geq \frac{1}{2}-\frac{n}{K(n)}\int_{0}^{1}(e^x(1-x))^n,dx\geq 0.$$ – Jack D'Aurizio Oct 03 '15 at 10:02
  • That proves that the difference between $\frac{1}{2}$ and the other piece goes to zero as $n\to +\infty$. – Jack D'Aurizio Oct 03 '15 at 10:04
  • @Jack I made a mistake rewriting the original inequality. Rewriting it correctly I get: $$-2/3\leq\big(\frac{n}{K(n)}\int_0^1(e^x(1-x))^n dx-\frac 1 2\big)K(n)\leq -1/2$$. Laplace method shows that the above expression goes to zero, however, because it kills the $1/2$ part correctly, but the error term is $O(1/n)$ while K(n)\sim n^{1/2}. – A.S. Oct 03 '15 at 11:52
  • @A.S. : I am really sorry, $(1-x^2)$ was definitely the wrong approximation. Answer fixed. – Jack D'Aurizio Oct 03 '15 at 11:56
  • @Jack It's better since you get the correction of the right order, but the constant is off and the upper bound on the integral needs to have a negative correction of order $O(1/n)$ as well. We need $$-\frac{2}{3n}<\int_{0}^{1}\left(e^t(1-t)\right)^n,dt - \sqrt{\frac{\pi}{2n}} < -\frac{1}{2n}$$ – A.S. Oct 03 '15 at 12:25
1

I only got asymptotic inequality. Numerical exploration tells us that the quantity in question - properly rescaled monotonically approaches the left limit in the inequality.

Asymptotics. After expanding $e^n=\sum_{k=0}^{n}\frac{n^k}{k!}+\frac1{n!}\int_0^ne^t(n-t)^ndt$, making a substitution $t=nx$ in the integral and rearranging the terms, the original inequality becomes

$$-\frac{2}{3}<\big(\int_{0}^{1}(e^t(1-t))^ndt -{\frac{K(n)}{2n}}\big)n < -\frac{1}{2}\tag{1}$$

where $K(n)=\displaystyle\frac{n!e^n}{n^n}=\sqrt{2\pi n}(1+\frac 1 {12n}+O(\frac 1 {n^2})\big)\tag{2}$

Use Laplace method to evaluate $\displaystyle I(n)=\int_{0}^{1}\big(e^t(1-t))^ndt=\int_0^1e^{nh(t)}dt$ where $$h(t)=t+\log(1-t)$$ $h$ achieves its maximum at $t=0$ and takes on values $h(0)=0, h'(0)=0,h^{(i)}(0)=-(i-1)!=h_i$ for $i\geq 2$. Expand $h(x)=h_0+\sum_{i\geq 2}h_ix^i/i!$ to get

Applying Laplace's method we get $$I(n)=e^{nh_0}\Big(\frac 1 2\frac {\sqrt{2\pi \sigma}} {n^{1/2}}+\frac {h_3\sigma^2}{3n}+\frac{\sqrt{2\pi \sigma}}{n^{3/2}}\big(\frac{h_4\sigma^2}{16}+\frac{5h_3^2\sigma^3}{48}\big)+\big({h_3h_4\sigma^4\over 3}+{8h_5\sigma^3 \over 5}\big)\frac {1} {n^2}+O(\frac 1 {n^{5/2}})\Big)=\frac{\sqrt{2\pi}}{2n^{1/2}}-\frac 2 {3n}+\frac {\sqrt{2\pi}} {24n^{3/2}}+{12\over 5n^2}+O(\frac 1 {n^3})\tag{3}$$

where $\sigma=|1/h_2|$. Note that in our case the the maximum is achieved at the boundary, hence compared to the case when maximum is achieved in the middle we get a factor of $1/2$ in from of even powers of $x$ in Taylor the expansion before integration and odd powers of $x$ aren't cancelled out.

Plug $(3)$ into (1) using (2) to get $$-\frac 2 3<-\frac 2 3+{12\over 5n}+O(\frac 1 {n^2})<-\frac 1 2$$ which holds.

For small $n$, values of $G(n)=\big(I(n)-\frac{K(n)}{2n}\big)n$ are $G(1)=e/2-2\approx -0.64, G(2)=1/(e^2-10)\approx -0.65274, G(3)=1/9(e^3-26)\approx -0.65716$

so it looks like the G(n) is monotonically decreasing to $-2/3$ which we know happens asymptotically and is already very close.

Can you prove inequality for all $n$?

A.S.
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