The nth term is a[n]=a[n-1]+10^-(n+1) and a[0]=1.1.
The next 3 terms are 1.11111 1.111111 1.1111111.
As n approaches infinity the nth term is 10/9 as a decimal.
The series is not a sum, but each term is a sum in itself of 1+(the sum of)10^-(r+1) [for 0<=r<=n], a geometric series with sum to n terms=10/9*(1-10^-(n+1)).