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

For $j,n\in\mathbb Z_+$, let $$ L_{j,n}^{(t)}= \sum_{m=0}^{n} \Bigl(-\frac 12\Bigr)^{n-m}{n\choose m}{m+j+1\choose m+1} \left( \frac {1}{t+\frac 12}\right)^{m+j+2} $$ and $$ L_{j,n} =\sum_{t=1}^\infty L_{j,n}^{(t)} $$

We would like to show that for $j,n\ge 0$ $$ (*)\quad \quad L_{2j,2n}>0. $$ Hypothesis (*) is supported by the numerics we did. In fact, it seems that $L_{j,n}>0$ is positive for all $j>0$ and all $n$. However, we only need the result for even indexes.

What do we know? Namely, the following easy facts:

  • $L_{j,0}>0$ for all $j$.
  • $L_{0,2n}>0$ for all $n$.
  • $L_{j,n}^{(1)}>0$ for all $n$ and $j>0$.

These easy facts follow from the observation that

$$ L_{j,n}^{(t)}=\frac {(-1)^j}{j!}\frac {d^j}{dz^j} \left( \frac {1}{z^2} \left[ \frac {1}{z}-\frac 12\right]^n \right)\left|_{z=t+\frac 12}\right. $$

In general, we need to estimate the sign of $$ \sum_{t=1}^\infty \Phi_n^{(j)}(t+\frac 12),\quad\text{where } \Phi_n(z)=\frac {1}{z^2} \Bigl[ \frac {1}{z}-\frac 12\Bigr]^n. $$ Heuristically, $$ (-1)^j j! L_{j,n}=\sum_{t=1}^\infty \Phi_n^{(j)}(t+\frac 12) \approx -\Phi_n^{(j-1)}(\frac 32) $$ and the sign of $\Phi_n^{(j)}(z)$ is $(-1)^j$ for every $z<2$, which supports the conjecture as well.

share|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.