Calculation shows that the sum is less than 1/2 for 3 < P(n) < 17 or 2 < n < 7:
5 Part2=1
10 for N=1 to 6
20 Part1=Part1+1/prm(N)^2
30 Part2=Part2/prm(N)
40 Tot=Part1+Part2:print N,Tot
50 next
n sum of terms
1 0.75
2 0.5277777777777777776
3 0.4344444444444444443
4 0.4262811791383219953
5 0.4302166376192350217
6 0.4357341969829482315
After that:
The Basel problem's solution states that the sum of the reciprocals of the squares of all the positive integers is pi^2/6 ~= 1.6449340668482264362. The sum of the first 16 of these terms is 1.5843465334449870744, leaving 0.0605875334032393618 for the remaining infinity of integers, that is, Sum{17 to infinity}1/k^2 ~= 0.0605875334032393618.
The sum of the reciprocals of the squares of the primes up to 16 is 1/4+1/9+1/25+1/49+1/121+1/169 ~= 0.4357008969496481982.
1/(2*3*5*7*11*13) ~= 0.0000333000333000333
Thus, the partial sum 1/2^2 + ... + 1/13^2 + 1/(2*3*5*7*11*13) plus the sum of the squares of the reciprocals of all the integers after 16, adds up to only .4963217303861876. For higher values of n, the term corresponding to 1/(2*3*5*7*11*13) only decreases, and terms without prime squares in the denominator are excluded from the previous part, the total will be less than .4963217303861876, and all the more so, less than 1/2.
We might be interested in knowing the limit of the total is as n gets higher.
In the table below, we introduce m, which equals P(n) when n is prime but is just a sequential integer otherwise.
Again we use the sum of the reciprocals of all the squares after m as an upper bound of the error when using a shortened series, when n is sufficiently high.
The result for m=1 is meaningless as 1 is not a prime and no primes have been counted yet.
The value of the sum of course changes only for prime m, while the sum of all following reciprocals of squares of all integers changes at each entry.
m value of sum sum of reciprocals of all squares
after m
1 1 0.6449340668482264362
2 0.75 0.3949340668482264362
3 0.5277777777777777776 0.2838229557371153251
4 0.5277777777777777776 0.2213229557371153251
5 0.4344444444444444443 0.1813229557371153252
6 0.4344444444444444443 0.1535451779593375474
7 0.4262811791383219953 0.133137014694031425
8 0.4262811791383219953 0.117512014694031425
9 0.4262811791383219953 0.105166335681685746
10 0.4262811791383219953 0.095166335681685746
11 0.4302166376192350217 0.0869018728717683907
12 0.4302166376192350217 0.0799574284273239462
13 0.4357341969829482315 0.0740402686640103368
14 0.4357341969829482315 0.0689382278476838062
15 0.4357341969829482315 0.0644937834032393618
16 0.4357341969829482315 0.0605875334032393618
17 0.4391630633875931828 0.0571273257907826144
18 0.4391630633875931828 0.0540409060376961946
19 0.4419312907606763486 0.0512708229352031199
20 0.4419312907606763486 0.0487708229352031199
21 0.4419312907606763486 0.0465032492390579952
22 0.4419312907606763486 0.0444371335365786564
23 0.4438215513152781745 0.0425467743683366905
24 0.4438215513152781745 0.0408106632572255794
25 0.4438215513152781745 0.0392106632572255794
26 0.4438215513152781745 0.0377313733163971771
27 0.4438215513152781745 0.0363596312039143239
28 0.4438215513152781745 0.0350841209998326913
29 0.445010607629499567 0.0338950603577399446
30 0.445010607629499567 0.0327839492466288335
31 0.4460511902062455028 0.0317433665203020906
32 0.4460511902062455028 0.0307668040203020906
33 0.4460511902062455028 0.0298485303747557177
34 0.4460511902062455028 0.0289834784716415309
35 0.4460511902062455028 0.028167151941029286
36 0.4460511902062455028 0.0273955470027576812
37 0.4467816503913138827 0.0266650868128380318
38 0.4467816503913138827 0.0259725660372147632
39 0.4467816503913138827 0.0253151038412910288
40 0.4467816503913138827 0.0246901038412910288
41 0.447376534388802876 0.0240952198436705649
42 0.447376534388802876 0.0235283264196342837
43 0.4479173672714389302 0.0229874935369950193
44 0.4479173672714389302 0.0224709646113751846
45 0.4479173672714389302 0.0219771374508813575
46 0.4479173672714389302 0.021504547658820866
47 0.4483700607979214267 0.0210518541323382947
. . .
2931154 0.4522473985006147804 0.000000341162580343
So the estimate when m has reached 2931154 is 0.4522473985006147804 and the highest it could possibly be is no more than this value plus 0.000000341162580343.
from
10 Pi=atan(1)*4
20 N=1:Sumfromn=Pi*Pi/6
30 Primepart1=0:Primepart2=1
40 while N<48
50 if N>1 and prmdiv(N)=N then Primepart1=Primepart1+1/(N*N):Primepart2=Primepart2/N
60 Tot=Primepart1+Primepart2
70 Sumfromn=Sumfromn-1/(N*N) ' sum starting at the next n using all integers
80 ' so it's the possible error on this n
90 print N,Tot,Sumfromn
100 N=N+1
110 wend
From the internet, it's reported that the actual limit is 0.4522474200410654985...
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Posted by Charlie
on 2013-06-22 16:06:17 |
computer assisted solution