Question

Let $a_n=\sum _{k=1}^n\frac{1}{k(n+1-k)}$, then for $n\ge 2$

• A. $a_{n+1}>a_n$
• B. $a_{n+1}<a_n$
• C. $a_{n+1}=a_n$
• D. $a_{n+1}-a_n=1/n$

Answer 1

Answer: (A)

$a_{n+1}>a_n$

$a_n={\sum }_{k=1}^n[\frac{1}{k}+\frac{1}{(n+1-k)}]\frac{1}{(n+1)}{a}_n=\frac{1}{(n+1)}[{\sum }_{k=1}^n\frac{1}{k}+{\sum }_{k=1}^n\frac{1}{n+1-k}]{\sum }_{k=1}^n\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}{\sum }_{k=1}^n\frac{1}{n+1-k}=\frac{1}{n}+\frac{1}{n-1}+....+\frac{1}{3}+\frac{1}{2}+\frac{1}{1}$

Hence, ${\sum }_{k=1}^n\frac{1}{k}={\sum }_{k=1}^n\frac{1}{n+1-k}$
$a_n=\frac{1}{(n+1)}[{\sum }_{k=1}^n\frac{1}{k}+{\sum }_{k=1}^n\frac{1}{n+1-k}]=\frac{2}{(n+1)}{\sum }_{k=1}^n\frac{1}{k}{a}_{n+1}={\sum }_{k=1}^{n+1}\frac{1}{k(n+2-k)}=\frac{1}{(n+2)}{\sum }_{k=1}^{n+1}(\frac{1}{k}+\frac{1}{(n+2-k)}){\sum }_{k=1}^{n+1}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{n}+\frac{1}{n+1}{\sum }_{k=1}^{n+1}\frac{1}{(n+2-k)}=\frac{1}{n+1}+\frac{1}{n}+\frac{1}{n-1}+....+\frac{1}{3}+\frac{1}{2}+\frac{1}{1}$

Hence, ${\sum }_{k=1}^{n+1}\frac{1}{k}={\sum }_{k=1}^{n+1}\frac{1}{(n+2-k)}$

$a_{n+1}=\frac{1}{(n+2)}({\sum }_{k=1}^{n+1}\frac{1}{k}+{\sum }_{k=1}^{n+1}\frac{1}{(n+2-k)})=\frac{2}{(n+2)}{\sum }_{k=1}^{n+1}\frac{1}{k}{a}_{n+1}=\frac{2}{(n+2)}{\sum }_{k=1}^{n+1}\frac{1}{k}+\frac{2}{(n+2)(n+1)}{a}_{n+1}=\frac{2}{(n+2)}\frac{(n+1)}{2}{a}_n+\frac{2}{(n+2)(n+1)}+\frac{(n+1)}{(n+2)}{a}_n+\frac{2}{(n+2)(n+1)}{a}_{n+1}>a_n$