For m>n consider:

|Sm - Sn| = |(Sm - Sm-1) + (Sm-1 - Sm-2) + ... + (Sn+1 - Sn)|

................<= |Sm - Sm-1| + |Sm-1 - Sm-2| + ... + |Sn+1 - Sn|

................< 2^-(m-1) + 2^-(m-2) + ... + 2^-n

.................= 2^-n { 1 + 2^-1 + 2^-2 + ... 2^-(m-n) }

.................= 2^-n (1 - 2^[-(m-n)+1]) / (1-2^-1)

..................< 2^(-n+1)

hence as m, n go to infinity |Sm-Sn| -> 0, that is (Sn) is a Cauchy

sequence and hence converges.

No, consider the partial sums of the harmonic series.(b) Is the result in (a) true if we only assume that |Sn+1 - Sn|<1/n for all n inN?

Sn = sum 1/r , r=1 .. n

We know that the harmonic series diverges hence (Sn) diverges, but:

|Sn+1 - Sn|= 1/(n+1) < 1/n.

Which provides the required counter example.

RonL