A lower bound for a sequence.

Question

Suppose that $n\geq3$. I want to prove that

$ v_n = \sum\limits_{p=1}^{n-1} \frac{1}{p(n+1-p)(n-p)} \geq \frac{2n+1}{2n^2} .$

Answer 1

By partial fraction decmposition (treating $p$ as $x$) $$\frac{1}{p(n+1-p)(n-p)}=\frac{1}{n(n+1)}\cdot\frac{1}{p}+\frac{1}{n}\cdot\frac{1}{n-p}-\frac{1}{n+1}\cdot\frac{1}{n+1-p}$$ hence by summing both sides for $p=1,2,\ldots,n-1$ we get:

$$\sum_{p=1}^{n-1}\frac{1}{p(n+1-p)(n-p)} = \frac{H_{n-1}}{n(n+1)}+\frac{H_{n-1}}{n}+\frac{1-H_n}{n+1}$$ or, by simplifying,

$$\sum_{p=1}^{n-1}\frac{1}{p(n+1-p)(n-p)} = \frac{2H_{n-1}+n-1}{n(n+1)}\geq\frac{n+2}{n(n+1)}\geq\frac{2n+1}{2n^2}.$$