Inequality with power function and binomial coefficients

Asked Aug 10 '15 at 04:04

Active Aug 10 '15 at 05:01

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Any suggestion on how to proceed to show:

$$\frac{2(m+1)^m -1 }{(m+1)m} - \sum_{k=0}^{m} {{m}\choose{k}} \frac{m^k}{(k+1)^2} >0 $$ where $m\geq 2$ is of course an integer.

Numerical results seem to confirm its validity.

edited Aug 10 '15 at 05:01

asked Aug 10 '15 at 04:04

Learner

That's an interesting problem. Where did you encounter it? – abiessu Aug 10 '15 at 04:07
1

$$ \left(\frac\partial{\partial q}q\right)^2\sum_{k=0}^m\binom mk\frac{q^k}{(k+1)^2}=\sum_{k=0}^m\binom mkq^k=(1+q)^m $$
might help.
– joriki Aug 10 '15 at 04:12
@abiessu This came up in some applied work. – Learner Aug 10 '15 at 21:22
@joriki Thanks for the suggestion. However, how to read $\left( \frac{\partial}{\partial q} q \right)^2$? – Learner Aug 10 '15 at 21:26
As $\frac\partial{\partial q}q\frac\partial{\partial q}q$. – joriki Aug 10 '15 at 21:55
@joriki Thanks. I feel this is the right direction but would appreciate more light on it. – Learner Aug 11 '15 at 00:45

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