Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to calculate, or bound from above, the following sum $$ \sum_{i=0}^n(n-2i)^p{p \choose i}, $$ here $p\geq 2$.

Any references are very welcome.

Thank you.

share|improve this question
    
Well, it is not bigger than $(2 n)^p$ –  Igor Rivin Apr 13 '12 at 16:35
2  
$p$ fixed, $n\to\infty$? $n$ fixed, $p\to\infty$? $n,p$ both going to infinity in some unspecified way? –  Gerry Myerson Apr 13 '12 at 23:14
1  
The paper info.tuwien.ac.at/panholzer/Papers/P13.pdf has some results that seem quite similar, so perhaps the methods used there can be adapted to Michael's problem. –  Richard Stanley Apr 28 '12 at 21:30
1  
But why crosspost under different names? Or am I missing something? –  Felix Goldberg Sep 18 '12 at 13:06
1  
Not relevant to your question, but it can be shown that all the zeros of the polynomial $\sum_{i=0}^p (x-2i)^p{p\choose i}$ have real part $p$. –  Richard Stanley Oct 24 '13 at 0:17

1 Answer 1

On of methods for finding an approximation or bound for a finite sum is using the following formula $$ \sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$$ and by taking $f(i)= (n-2i)^p\binom {p}{i}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

Moreover, the sharp bounds of Bernoulli numbers has been computed ,(see here ) So you can also try to find a bound for your sum.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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