Abandoned, incompolete; read carefully








Question: let  A  be the set of positive integers whose decimal expansions
do not contain (say) a 2 nor a 5. What is  S=\sum_{n \in A} (1/n)  ?

Answer: Clearly  A  is the disjoint union of its subsets  A_k  of such
numbers of exactly  k  digits, and  S = \sum_k S_k  where  the sums  S_k
are defined similarly. 

Expand  S_k  as

	Sum_{n in A_k}  1/n    =

    Sum_{m in A_{k-1}} (  1/(10n) + 1/(10n+1) + 1/(10n+3) + ... + 1/(10n+9) )

These eight terms sum to something less than  8/(10n); we may obtain a
lower bound (and more accurate expressions) by computing a Taylor series
in  z=1/n : the sum is 

	(8/10) (1/n) - (38/100) (1/n)^2 + (256/1000) (1/n)^3 - ...

This allows us to evaluate  S_k  as

	(8/10) S_{k-1} -(.38) Sum (1/n^2, n in A_{k-1}) + ...

More precisely,  S_k - (.8) S_{k-1} is negative and of magnitude bounded 
above by .38 Sum(1/n^2) . The sum is to be taken over  A_{k-1}  only, but we
may obtain a weaker upper bound by summing over _all_  (k-1)-digit
numbers. That sum is in turn less than the integral of  1/x^2  over the
range  10^(k-2) - 1  to  10^(k-1) - 2, which is  

	(.9) 10^(-(k-2)) + (.98) 10^(-2(k-2)) + ...

A tiny tightening of these inequalities, and a shift of index, gives

	.8 S_k - .38 * 9/10^k < S_{k+1} < .8 S_k.

By induction we then have  

	(.8)^n S_k - .38 * (9/10^k)( (.8)^n - (.1)^n)/(.7) < S_{k+n} < (.8)^n S_k

It is then simple to sum the geometric series to estimate
S_k + S_{k+1} + ...  : it's less than  5 S_k, but by no more than
19/10^k .

Thus our idea is simply to evaluate by hand the first few  S_k's, and
when we tire we declare the sum of all of them to be the sum of all
but the last, plus 5 times the last. Our answer will be too large but
by at most  (1/2) 10^( - (k-2) ), i.e., the answer is 
"correct to within  k-2  decimals".




ok:=proc(n) local k : if n=0 then true else 
 k:=n mod 10: if (k=2) or (k=5) then false else ok( (n-k)/10 ) fi:fi:end:

Digits:=30: # :-)
for k from 1 to 5 do
  S:=0: for i from 10^(k-1) to 10^k-1 do if ok(i) then S:=S+evalf(1/i) fi:od:
  S.k:=evalf(S):
od: print(seq(S.k, k=1..5));
2.12896825396825396825396825397
1.39055196196383020583358259504
1.09102512106214326702081773525 
 .87114105194401603041413667953
 .69677879363656275519104058389

for k to 5 do lprint(add(S.j,j=1..k-1) + 5*S.k) od:
10.6448412698412698412698412699
9.08172806378740499742188122917
8.97464582124280050919163952526
8.96625059671430759317905198192
8.96558035712105724747770818327


Now, we can improve our estimates a little. The set  A_k  contains only
N = 7*8^(k-1)  of the  k-digit numbers, so comparing to the smallest
k-digit numbers, the sum of  1/n^2  on this set will be less than
(N)/(10^(2(k-1))) = (175/2)(.08)^k, an improvement over our previous 
(.9)(.1)^k for large  k.
Thus the sum of all S's starting with  S_k  will differ from  5S_k  by
no more than  some multiple of (.08)^k; a good multiple is, empirically,
about 17.78 . This leads to an estimated value of the sum of all S's to be
8.96552