D(n) denotes the number of digits in the positive integer N.
Evaluate:
∞
Σ {(D(n))!}-1
n=1
Lets break up the sum by the number of digits.
There are 9 1-digit n, those terms sum to 9/1!
There are 90 2-digit n, those terms sum to 90/2!
In general there are 9*10^(k-1) k-digit n, those terms sum to 9*10^(k-1)/k!
Then the summation can be expressed as sum {k = 1 to inf} 9*10^(k-1)/k!
Factor 10/9 out. Then we have (10/9) * sum {k = 1 to inf} 10^k/k!
This last summation is the series expansion of e^x evaluated at 10, but starting at the second term.
Then we have (10/9)*(e^10-1) as the answer.
Solution
