`4)`
`a)` `u_n=1+1/n`
`=> u_{n+1}=1+1/{n+1}`
`b)` Ta có:
`u_{n+1}-u_n`
`=1+1/{n+1}-(1+1/n)`
`=1/{n+1}-1/n`
`∀n∈N`* ta có:
`n+1>n`
`⇒1/{n+1}<1/n`
`⇒1/{n+1}-1/n<0`
`⇒u_{n+1}-u_n<0`
Vậy `u_{n+1}<u_n ∀n∈N`*
`5)`
`a)` `v_n=5n-1`
`=> v_{n+1}=5(n+1)-1=5n+4`
`b)` Ta có:
`v_{n+1}-v_n=5n+4-(5n-1)=5>0`
`⇒v_{n+1}-v_n>0`
Vậy `v_{n+1}>v_n ∀n∈N`*