## Precalculus: Mathematics for Calculus, 7th Edition

in order a sequence to be arithmetic $a_{n+1}-a_{n}=a_{n-}a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$ So $2a_{n}=2\times \dfrac {1}{1+2n}$ $a_{n+1}+a_{n-1}=\dfrac {1}{1+2\left( n+1\right) }+\dfrac {1}{1+2\left( n-1\right) }=\dfrac {1}{2n+3}+\dfrac {1}{2n-1}=\dfrac {4n+2}{4n^{2}+4n-3}=\dfrac {2\left( 2n+1\right) }{4n^{2}+4n+1-4}=\dfrac {2\left( 2n+1\right) }{\left( 2n+1\right) ^{2}-4}=\dfrac {2}{\left( 2n+1\right) +\dfrac {4}{2n+1}}$ $\Rightarrow 2a_{n}\neq a_{n+1}+a_{n-1}$ So this sequence is not arithmetic