当n≥2时
Sn=lgn
Sn-1=lg(n-1)
Sn-Sn-1=an=lgn-lg(n-1)=lg[n/(n-1)]
当n=1时S1=a1=0不属于an=lg[n/(n-1)]
n=1时a1=0
n≥2时an=an=lg[n/(n-1)]求证：0<=an<=lg2n=1时an=0n≥2时an=lg[n/(n-1)] =lg(1+(1/(n-1)))0<1/(n-1)≤11<1+1/(n-1))≤20=lg1≤lg[n/(n-1)] ≤lg2∴0≤an≤lg2