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!

  • 1
    $\begingroup$ May be ther should be $\left(\frac{p}{1-p}\right)^i$ $\endgroup$ – Norbert Apr 23 '13 at 18:16

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 $q<1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.