`1/5+1/12+1/24+1/32+1/64+1/72+1/81`

`=1/5+(1/12+1/24+1/32)+(1/64+1/72+1/81)`

Ta có:

`1/5 ` giữ nguyên

`1/12+1/24+1/32<1/12+1/12+1/12=1/(12).3`

`1/64+1/72+1/81<1/60+1/60+1/60=1/(60).3`

`⇒1/5+1/12+1/24+1/32+1/64+1/72+1/81<1/5+1/(12).3+1/(60).3`

`⇒1/5+1/12+1/24+1/32+1/64+1/72+1/81<1/5+1/4+1/20`

`⇒1/5+1/12+1/24+1/32+1/64+1/72+1/81<4/20+5/20+1/20=1/2`

`⇒1/5+1/12+1/24+1/32+1/64+1/72+1/81<1/2`

Vậy `1/5+1/12+1/24+1/32+1/64+1/72+1/81<1/2`(đpcm)