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.
The inequality you suggest in the comments:$_2F_1(n+\alpha,1;n;-\lambda)\geq\frac{1}{1+\lambda(1+\alpha/n)}$ is not true. Take $n=2, a=1,\lambda=-0.1$. Then the $_2F_1(n+\alpha,1;n;-\lambda)=105/121$. On the other hand $1/(1+\lambda(1+\alpha/n))=20/23$.
It is clear that $20/23>105/121$ and hence the inequality does not hold.
On then other hand you could take the first two terms from the series representations of the hypergometric function and very easily see that:$$ _2F_1(n+\alpha,1;n;-\lambda)\geq1-\lambda(1+\alpha/n) $$