Ta có: `u_n=n/{3^n}` $(n\in N^*)$
`=>u_{n+1}={n+1}/{3^{n+1}}`
`=>{u_{n+1}}/{u_n}={n+1}/{3^{n+1}} : n/{3^n}`
$=\dfrac{(n+1). 3^n}{n. 3^{n+1}}=\dfrac{(n+1).3^n}{n.3. 3^n}$
`={n+1}/{3n}=n/{3n}+1/{3n}=1/ 3 +1/ {3n}`
`\forall n\in N`* ta có:
`\qquad n\ge 1`
`=>3n\ge 3`
`=>1/{3n}\le 1/ 3`
`=>1/ 3 +1/{3n}\le 1/ 3 + 1/ 3=2/ 3`
`=>{u_{n+1}}/{u_n}\le 2/ 3`
Vậy `{u_{n+1}}/{u_n}\le 2/ 3` $\forall n\in N$* $(đpcm)$
