a(1)<=...<=a(n) Adding i to the i-th member we have a strictly increasing series. We work with strictly increasing series. So we compute a(1) ^ 2 + a(2) ^4 + a(3) ^ 8 + .... But we need a lemma: if a<=b natural numbers, then a,b can be gained (computed) back from f(a,b)=a+b^2. Indeed, obviously b^2 <= f(a,b) = a + b^2 <= b + b^2 < 1 + 2*b + b^2 = (b+1) ^2 So (b+1)^2 is the smallest square number which is bigger than f(a,b), so we can compute back b and thus a. Now if we compute the above expression up to the member a(k), we obtain f(k) = a(1) ^ 2 + a(2) ^4 + a(3) ^ 8 + ....+ a(k) ^ (2^k) The member whose square is to be added to come to f(k+1) is a(k+1)^(2^k). We can apply our lemma, if f(k) is not greater than the above number. Let a(k) be A, then all a(i) up to it will be be <= A, f(k) <= A + A^2 + A^4 + ... + A^(2^k) Even the right hand side expression is <= a(k+1) ^ (2^k) , because a(k+1) must be at least (A+1), and if we apply the binomial expansion to (A+1)^(2^k) we have more more and bigger members than enough.