How to Prove This Complex Inequality Involving Factorials and Sums?

[klim]

Hallo, can someone help me to proof this inequality:

\(\displaystyle (\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^m \frac{\lambda^j}{j!} (m+1-j) \leq \frac{\lambda}{\lambda-m} \)

with condition \(\displaystyle m+1 < \lambda \).

\(\displaystyle \lambda\) is real und \(\displaystyle m\) is integer.

 

[Jhoneine]

Proof:Let f(m) = (\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^m \frac{\lambda^j}{j!} (m+1-j).We want to show that f(m) ≤ \frac{\lambda}{\lambda-m} for m+1 < \lambda.Since f(m) is a continuous function, we can use the Mean Value Theorem to show that there exists a c ∈ (m,m+1) such thatf'(c) = \frac{f(m+1)-f(m)}{m+1-m} = \frac{f(m+1)-f(m)}{1}.Butf(m+1) = (\lambda-(m+2)) \cdot \frac{(m+1)!}{\lambda^{m+2}} \cdot \sum_{j=0}^{m+1} \frac{\lambda^j}{j!} (m+2-j)andf(m) = (\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^m \frac{\lambda^j}{j!} (m+1-j).Substituting in f'(c) gives:f'(c) = \frac{(\lambda-(m+2)) \cdot \frac{(m+1)!}{\lambda^{m+2}} \cdot \sum_{j=0}^{m+1} \frac{\lambda^j}{j!} (m+2-j) - (\lambda-(m+1)) \cdot \frac{m!}{\lambda^{m+1}} \cdot \sum_{j=0}^m \frac{\lambda^j}{j!} (m+1-j)}{1}.Simplifying this expression gives:f'(c) = \frac{(\lambda-(m+1)) \cdot \frac{(m+1)!}{\lambda^{m+2}} \