# If $a_{n}=\sum_{k=1}^{n}\;\large\frac{1}{k(n+1-k)}$, then for $n\;\geq\;2$

$(a)\;a_{n+1} > a_{n}\qquad(b)\;a_{n+1} = a_{n}\qquad(c)\;a_{n+1} < a_{n}\qquad(d)\;a_{n+1} - a_{n}=n$

Answer : (c) $\;a_{n+1} < a_{n}$
Explanation : $a_{n}=\large\frac{1}{n+1}\;\sum_{k=1}^{n}\;(\large\frac{1}{k}+\large\frac{1}{n+1-k})$
$=\large\frac{2}{n+1}\;\sum_{k=1}^{n}\;\large\frac{1}{k}$
$For \; n \geq 2$
$\large\frac{1}{2}\;(a_{n}-a_{n+1})=\large\frac{1}{n+1}\sum_{k=1}^{n}\;\large\frac{1}{k}-\large\frac{1}{n+2}\sum_{k=1}^{n+1}\;\large\frac{1}{k}$
$=(\large\frac{1}{n+1}-\large\frac{1}{n+2})\;\sum_{k=1}^{n}\;\large\frac{1}{k}-\large\frac{1}{(n+1)(n+2)}$
$=\large\frac{1}{(n+1)(n+2)}\;\sum_{k=2}^{n}\;\large\frac{1}{k}\; > 0$
$=a_{n} > a_{n+1}$
$=a_{n} < a_{n+1}\;.$