Recent content by klim

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...

Diplom-Thesis!

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}...

is there any adaquate bound for Incomplete Gamma Function?

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...

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.

Hallo, can someone help me to proof this inequality: $$ (\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^m \frac{\lambda^j}{j!} (m+1-j) \leq \frac{\lambda}{\lambda-m} $$ with condition $$ m+1 < \lambda $$. $$\lambda$$ is real und $$m$$ is integer.

Hallo, can someone help me to proof this inequality: $$ 1-(\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^{m} \frac{\lambda^j}{j!} \leq \frac{\lambda}{(\lambda-(m+1))^2} $$ under condition $$ m+1 < \lambda $$.