I am stuck on proving the inequality in
$LHS:=\sum_{k=1}^{\infty}(-\lambda)^k\prod_{i=1}^k \left(1+\frac{\alpha}{n+i}\right)\geq \sum_{k=1}^{\infty}(-\lambda)^k\left(1+\frac{\alpha}{n}\right)^k=\frac{-\lambda(\alpha+n)}{n+\lambda(\alpha+n)}:=RHS$
where $\lambda, \alpha, n\geq 0$, and $\lambda\left(1+\frac{\alpha}{n}\right)<1$ which ensures convergence of RHS. If necessary, one may assume $2\alpha$ is an integer.
Obviously, LHS=RHS if $\alpha=0$ or $n\rightarrow\infty$. Numerical evaluations indicate (no proof):
- The inequality is valid, also if $n$ at the right hand side is replaced by $n+\frac{1}{2}$
- The difference LHS-RHS increases monotonically in $a$ and decreases monotonically in $n$
Background (no need to read this): If $S$ is a Gamma distribution with shape $\alpha$ and scale $\lambda$, and $n$ an even integer, then $LHS = \frac{1}{x_n}\sum_{k=n+1}^{\infty}x_k$ where $x_k = \frac{E[(-S)^k]}{k!} = \frac{(-\lambda)^k\Gamma(\alpha+k)}{k!\Gamma(\alpha)} $. Thus, LHS is the scaled remainder when $\sum_{k=0}^{\infty}x_k=E[\exp(-S)]=\frac{\alpha\lambda}{(1+\lambda)^{\alpha+1}}$ is truncated after $n$ terms.