MHB Can this inequality be proven under given conditions?

[klim]

Hallo, could comeone help me to proof this inequality:

$$ \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{2}{\sqrt{\lambda}} $$.

under condition $$ m+1 < \lambda $$.

$$\lambda$$ is real and $$m$$ is integer.

 

[Fernando Revilla]

$$ \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{2}{\sqrt{\lambda}} $$. under condition $$ m+1 < \lambda $$. $$\lambda$$ is real and $$m$$ is integer.

I suppose you mean $\lambda>0$ and $m \geq 0,$ otherwise the inequality has no sense. Use induction. For the inductive step, $$ \frac{(m+1)!}{\lambda^{m+2}} \cdot \sum_{j=0}^{m+1}\frac{\lambda^j}{j!}=\frac{m!}{\lambda^{m+1}}\cdot\frac{m+1}{\lambda}\left(\sum_{j=0}^{m}\frac{\lambda^j}{j!}+\frac{\lambda^{m+1}}{(m+1)!}\right)\\
\underbrace{\leq}_{\text{Hip. induc.} }\frac{m+1}{\lambda}\cdot \frac{2}{\sqrt{\lambda}}+\frac{(m+1)!}{\lambda^{m+2}}\cdot \frac{\lambda^{m+1}}{(m+1)!} $$ $$=\frac{m+1}{\lambda}\cdot \frac{2}{\sqrt{\lambda}}+\frac{1}{\lambda}\underbrace{\leq}_{\text{if }{}1+m<\lambda-1}\frac{\lambda-1}{\lambda}\cdot \frac{2}{\sqrt{\lambda}}+\frac{1}{\lambda}$$ $$=\frac{2}{\sqrt{\lambda}}\left(1-\frac{1}{\lambda}+\frac{1}{2\sqrt{\lambda}}\right).$$ Now, consider $f: (0,+\infty)\to \mathbb{R},\; f(\lambda)=1-\dfrac{1}{\lambda}+\dfrac{1}{2\sqrt{\lambda}}.$ It is easy to prove that $f(\lambda)\leq \color{red}17/16$ for all $\lambda>0.$ As a consequence $$\frac{(m+1)!}{\lambda^{m+2}} \cdot \sum_{j=0}^{m+1}\frac{\lambda^j}{j!}\leq \frac{2}{\sqrt{\lambda}}\color{red}\cdot \frac{17}{16}\color{black}.\qquad\blacksquare$$

 

[klim]

Fernando Revilla,
thank you very much for your answer . But your solution has an error an the end.
$f(\lambda)=1-\dfrac{1}{\lambda}+\dfrac{1}{2\sqrt{\lambda}}$ is not always less OR equal ONE. For example: $\lambda=9$.
$f(\lambda)=f(9)=1-\frac{1}{9}+\frac{1}{2 \cdot 3}=1-\frac{1}{9}+\frac{1}{6}=\frac{19}{18} > 1$.
So, I think the last step is not correct!

 

[Fernando Revilla]

But your solution has an error an the end.
$f(\lambda)=1-\dfrac{1}{\lambda}+\dfrac{1}{2\sqrt{\lambda}}$ is not always less OR equal ONE. For example: $\lambda=9$.
$f(\lambda)=f(9)=1-\frac{1}{9}+\frac{1}{2 \cdot 3}=1-\frac{1}{9}+\frac{1}{6}=\frac{19}{18} > 1$.
So, I think the last step is not correct!

Right, I had plotted $f(\lambda)=1-\dfrac{1}{\lambda}-\dfrac{1}{2\sqrt{\lambda}}$ instead. I'll try to find a better bound.

 

[Fernando Revilla]

I can't get an adequate bound using the induction method. Have you covered the Incomplete Gamma Function? Then, you can try using $$\frac{m!}{\lambda^{m+1}}\sum_{j=0}^m\frac{\lambda^j}{j!}=\frac{e^{\lambda}\;\Gamma(m+1,\lambda)}{\lambda^{m+1}}.$$

 

[klim]

I can't get an adequate bound using the induction method. Have you covered the Incomplete Gamma Function? Then, you can try using $$\frac{m!}{\lambda^{m+1}}\sum_{j=0}^m\frac{\lambda^j}{j!}=\frac{e^{\lambda}\;\Gamma(m+1,\lambda)}{\lambda^{m+1}}.$$

is there any adaquate bound for Incomplete Gamma Function?

 

[Fernando Revilla]

is there any adaquate bound for Incomplete Gamma Function?

No, there isn't. But your problem is equivalent to prove that $$e^{\lambda}\;\Gamma(m+1,\lambda)\leq \frac{2\lambda^{m+1}}{\sqrt{\lambda}}\quad (m+1<\lambda).$$ Use the recurrence relation $\Gamma (a+1,x)=a\Gamma (a,x)+x^ae^{-x}$ to prove that $$e^{\lambda}\;\Gamma(m+2,\lambda)\leq \frac{2\lambda^{m+2}}{\sqrt{\lambda}}\quad (m+2<\lambda).$$

 

[klim]

I have another idea, how to solve this.

We multiply both sides by $\sqrt{\lambda}$ and our aim now is to proof:
$ m! \cdot \sum_{j=0}^{m} \frac{\lambda^{-(m+\frac{1}{2}-j)}}{j!} \leq 2 $.
At the next step we define the function $f(x)=m! \cdot \sum_{j=0}^{m} \frac{x^{-(m+\frac{1}{2}-j)}}{j!} $ with $x > 0$.
It's clear, that derivative $f'(x) < 0$ for any [math]x > 0[/math]. So, to proof our initial inequality, it's enough to proof, that $f(m+1) \leq 2$.
For x=m+1 we have
$f(m+1)=m! \cdot \sum_{j=0}^{m} \frac{(m+1)^{-(m+\frac{1}{2}-j)}}{j!}=\sum_{j=0}^{m} \frac{m!}{j! \cdot (m+1)^{(m+\frac{1}{2}-j)}}$.
But: Is it possible to proof: [math]f(m+1) \leq 2[/math] for any integer $m \geq 0$ ?

 

[Fernando Revilla]

Before reviewing your proposal, a previous question. What is the context of your problem?, examination? research?, ...

 

[klim]

Before reviewing your proposal, a previous question. What is the context of your problem?, examination? research?, ...

Diplom-Thesis!

 

[Fernando Revilla]

Diplom-Thesis!

Well, then surely is not an isolated problem. Perhaps you need it to prove another thing, perhaps the inequality is only a conjecture, etc. I haven't been able to prove it. Even using the Incomplete Gamma Function, I can't avoid the $17/16$ bound.

I'll try again, but I can't assure you anything.

 

[klim]

Well, then surely is not an isolated problem. Perhaps you need it to prove another thing, perhaps the inequality is only a conjecture, etc. I haven't been able to prove it. Even using the Incomplete Gamma Function, I can't avoid the $17/16$ bound.

I'll try again, but I can't assure you anything.

Right, it isn't an isolated problem. I have to proof one quite big theorem and a part of this proof is this inequality.

I have another idea, how could this theorem be proved.
We have to proof, that $ \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{2}{\sqrt \lambda} $ для $m+1 < \lambda $.

So we have:
$ \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{2}{\sqrt \lambda} \Leftrightarrow m! \cdot \sum_{j=0}^{m} \frac{\lambda^{j-m-\frac{1}{2}}}{j!} \leq 2 \Leftrightarrow m! \cdot \sum_{j=0}^{m} \frac{1}{\lambda^{m+\frac{1}{2}-j} \cdot j!} \leq 2 $.

Because of $ m+1 < \lambda $, it's enough to proof, that $ m! \cdot \sum_{j=0}^{m} \frac{1}{(m+1)^{m+\frac{1}{2}-j} \cdot j!} \leq 2 $.

Further: $ m! \cdot \sum_{j=0}^{m} \frac{(m+1)^{j-m-\frac{1}{2}}}{j!} = \frac{m!}{(m+1)^{m+\frac{1}{2}}} \cdot \sum_{j=0}^{m} \frac{(m+1)^j}{j!}=\frac{m!}{(m+1)^{m+\frac{1}{2}}} \cdot e^{m+1} \cdot \sum_{j=0}^{m} e^{-(m+1)} \frac{(m+1)^j}{j!}$
$=\frac{m!}{(m+1)^{m+\frac{1}{2}}} \cdot e^{m+1} \cdot Po(m+1)([0,m]) $ Using the identity theorem of power series, we can proof, that $Po(m+1)([0,m])=\int_{(m+1)}^{\infty} \left( e^{-t} \cdot \frac{t^m}{m!} \right) dt $

In this way we get: $ m! \cdot \sum_{j=0}^{m} \frac{(m+1)^{j-m-\frac{1}{2}}}{j!} = \frac{m!}{(m+1)^{m+\frac{1}{2}}} \cdot e^{m+1} \cdot \int_{(m+1)}^{\infty} \left( e^{-t} \cdot \frac{t^m}{m!} \right) dt $.

After many calculations we get:
$ m! \cdot \sum_{j=0}^{m} \frac{(m+1)^{j-m-\frac{1}{2}}}{j!}=\frac{m!}{(m+1)^{m+\frac{1}{2}}} \cdot e^{m+1} \cdot Po(m+1)([0,m]) \leq$
$\leq \sqrt{m+1} \cdot \int_{1}^{\infty} exp \left( (1-t)-\frac{m(1-t)^2}{2} \right) dt $.

And the actual question:

Is it possible to proof. that $ \sqrt{m+1} \cdot \int_{1}^{\infty} exp \left( (1-t)-\frac{m(1-t)^2}{2} \right) dt \leq 2 $

If it is possible, then the proof is done!