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How to show following inequality using Stirling approximation?$$\sum_{i=1}^n(\frac{p}{1-p})^n\cdot\frac{1}{(n+i)!(n-i)!} \leq \frac{1-p}{1-2p}$$ Any kind of hint will be appreciated. Thanks in advance!

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    $\begingroup$ May be ther should be $\left(\frac{p}{1-p}\right)^i$ $\endgroup$ – Norbert Apr 23 '13 at 18:16
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Hint. Set $q=\dfrac{p}{1-p}$. What we must show is equivalent to $$\sum_{i=1}^n q^i\cdot\frac{1}{(n+i)!(n-i)!}\le\frac1{1-q}.$$

This looks much easier, since we know $\displaystyle\sum_{i=0}^{\infty} q^i=\dfrac1{1-q}$ for $|q|<1$.

I wonder if the original problem sets any constraint on $p$ (or equivalently, $q$). For example, the above inequality obviously does not hold for $q<1$.

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