# sum alternating series inequality

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.

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The LHS is a hypergeometric series, and using Euler's representation for the Gauss's hypergeometric function one has: $$\mathrm{LHS} = -\lambda (n+a+1) \int_0^1 (1-t)^{n} \left(1 + \lambda t \right)^{-n-a-2} \mathrm{d} t$$ Maybe this could be useful. –  Sasha Aug 6 '12 at 20:39
Adding 1 to LHS and RHS, and replacing $a$ by $\alpha$ to avoid confusion then gives –  Alexander Aug 7 '12 at 7:11
$$_2F_1(n+\alpha,1;n;-\lambda)\geq\frac{1}{1+\lambda(1+\alpha/n)}$$ I still do not see how to prove this inequality. Proving that the derivative of the difference $LHS-RHS$ to $\alpha$ is positive would be sufficient. –  Alexander Aug 7 '12 at 7:19
After some rewriting, it turns out that I need to prove $$\int_0^1 \frac{t^{n-1}}{(1+\lambda(1-t))^{n+\alpha+1}}dt\geq\int_0^1 t^{\lambda(n+\alpha)+n-1}dt\,.$$ Can anybody help me? –  Alexander Aug 10 '12 at 7:51
your inequality is $f(t)=a-b>0$ for which I took the second derivative and saw that it was all positive. This means it was convex and increasing in $t$. The problem was that the closed form expression wasso ugly. I did it in MATLAB. –  Seyhmus Güngören Aug 10 '12 at 23:03

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:$$_2F_1(n+\alpha,1;n;-\lambda)\geq1-\lambda(1+\alpha/n)$$