20210820, 22:54  #23  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3^{2}×239 Posts 
Quote:
https://www.wolframalpha.com/input/?...%5En+converges 

20210820, 23:13  #24  
Apr 2020
1F4_{16} Posts 
Quote:
Last fiddled with by charybdis on 20210820 at 23:18 

20210820, 23:24  #25 
"Rashid Naimi"
Oct 2015
Remote to Here/There
3^{2}·239 Posts 
Well, from that we could jump to
sum_(n=1)^∞ (a/(a+1))^n = a https://www.wolframalpha.com/input/?...%2B1%29%29%5En and 1/x=a/(a+1) solve for x => x = 1/a + 1 and a!=0 and a + 1!=0 https://www.wolframalpha.com/input/?...29+solve+for+x So for the infinite sum: sum_(n=1)^∞ (1/a)^n as long as a is greater than 1, the sum will converge. So putting that in perceptive of the OP, the sum would likely convege. I think the smoke from my ears just set out the smole detectors. Thank you charybdis for your replies and humoring me. I do appreciate it very much. Last fiddled with by a1call on 20210820 at 23:24 
20210821, 00:12  #26  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
3^{2}·643 Posts 
Quote:
Let r_{i}=1/R_{i}. r is then always less than 1, and > 0. Sum of a(1/R)^{n} =a r^{n} converges for R > 1, 1 < r < 1, for a finite or infinite geometric series, n=1, 2, 3, .... (Any decent math textbook above a certain level will contain a proof. See https://en.wikipedia.org/wiki/Geometric_series) So the sum of the reciprocals of the exponents of Mersenne primes does converge, must converge. It's an open question whether there are an infinite number of Mersenne primes. But if they were infinitely many, and were mostly to fall near the spacing indicated as average for the Lenstra & Pomerance conjecture, R~1.47576139704882, that leaves considerable room for modest fluctuation from term to term relative to that convergence requirement, such as occasionally to R~1.0001 which would still converge even if all were so closely spaced, and even more so for R>>1.0001 or R>1.476. The empirical evidence provided by the 51 known data points spread over ~10^{8} is consistent with the Lenstra & Pomerance conjecture being correct. The sum of reciprocals over the 51 actual is actually ~7% lower than we get by summing what the conjecture predicts, beginning with a=1/2. The sum over the next ~6 we might expect to find below p~10^{9} cannot be greater than 10^{7}, and so must fall far short of making up the difference. See post 4 this thread. So far the 50 observed ratios of successive exponents of Mersenne primes range ~1.011 to 4.1024. (43112609/42643801 to 521/127) 

20210821, 01:52  #27 
"Rashid Naimi"
Oct 2015
Remote to Here/There
100001100111_{2} Posts 
Ok, to elaborate further:
* The infinite series sum: sum_(n=1)^∞ (1/a)^n converges for all real numbers a such that a>1 https://www.wolframalpha.com/input/?...%5En+converges * There are infinite prime numbers * The sum of reciprocals of all prime numbers diverges * We can formulate a in the 1st statement infinitely smaller than any given real number greater than 1 ** In this way we can make sure that the 1st few/many/given/finitenumber=m addends of the series in 1st statement are larger than any given firstfinitenumber=m of the reciprocalsofallprimenumbers *** This entails that it is guaranteed that at some point beyond m the addends of the reciprocalsofallprimenumbers will overtake/getlargerthan the addends of the formulated infinite series * All this to say there absolutely can be no conclusion that can be based on comparing the OPsum's convergence with a converging sum of a geometric series Corrections/enlightenments/comments are as always appreciated. Thank you very much for your time. Last fiddled with by a1call on 20210821 at 01:55 
20210821, 11:28  #28 
May 2018
2×97 Posts 
Mersenne Constant
Let's introduce the term Mersenne constant for constants obtained from converging sums of reciprocals involving the Mersenne prime exponents.
The following notations could prove useful when dealing with such Mersenne constants:  M_{Σ1/p} for the sum of the reciprocals of Mersenne prime exponents;  M_{Σ1/π(p)} for the sum of the reciprocals of π(p) of Mersenne prime exponents;  M_{Σ±1/p} for the alternating sum of the reciprocals of Mersenne prime exponents;  M_{Σ±1/π(p)} for the alternating sum of the reciprocals of π(p) of Mersenne prime exponents;  M_{Σ1/(p[n+1]p[n])} for the sum of the reciprocals of the difference between consecutive Mersenne prime exponents;  M_{Σ1/(π(p[n+1])π(p[n]))} for the sum of the reciprocals of the difference between consecutive π(p) of Mersenne prime exponents;  M_{Σ±1/(p[n+1]p[n])} for the alternating sum of the reciprocals of the difference between consecutive Mersenne prime exponents;  M_{Σ±1/(π(p[n+1])π(p[n]))} for the alternating sum of the reciprocals of the difference between consecutive π(p) of Mersenne prime exponents; etc. 
20210821, 13:13  #29 
May 2018
C2_{16} Posts 
On the basis of the 51 known Mersenne prime exponents:
M_{Σ1/p} = 1.448181855... M_{Σ1/π(p)} = 2.813838890... M_{Σ±1/p} = 0.2697096036... M_{Σ±1/π(p)} = 0.6628596647... M_{Σ1/(p[n+1]p[n])} = 3.218584673... M_{Σ1/(π(p[n+1])π(p[n]))} = 7.055035487... M_{Σ±1/(p[n+1]p[n])} = 0.6684789674... M_{Σ±1/(π(p[n+1])π(p[n]))} = 0.9880482016... Last fiddled with by Dobri on 20210821 at 13:15 
20210821, 14:19  #30 
Feb 2017
Nowhere
2^{2}·29·43 Posts 
So far, the most reasonable approach IMO is to apply the best guess about Mersenne prime exponents (that they increase approximately geometrically with ratio R ~ 1.47576139704882). This gives the simple estimate for the "tail" of the series of reciprocals after a given exponent p, of 1/(p*(R1)), or about 2.1/p, so if S_{p} is the sum of reciprocals up to the Mersenne prime exponent p, and S is the sum over all Mersenne prime exponents,
S ~ S_{p} + 2.10189393/p. This thread has led me to the discovery of a feature of the calculator app on my computer: Instead of typing input and transcribing output one digit at a time, I can copypaste plain text directly into the calculator, and copy the number in the calculator's display, and then paste it into a text file. Given my proficiency at making typos, this may be the most significant result of this thread so far 
20210821, 15:50  #31  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
3^{2}×643 Posts 
Quote:
So, S ~ 1.4481818554213 + 2.10189393 / 82589933 ~ 1.448181880871059... (On Windows 7 or 10 calculator, one can copy/paste multiple numbers and operators and spaces in a single go, including from web pages or pdf documents: try 1.4481818554213 + 2.10189393 / 82589933= ) (Yay, both Mersenne prime & exponent & sequence #, post number! That's only happened, um, at # 3, 7 and here in this thread) Last fiddled with by kriesel on 20210821 at 16:08 

20210822, 20:22  #32  
May 2018
2·97 Posts 
Quote:
Let's also assume that the number of Mersenne primes is infinite. Corollary: M_{Σ1/π(p)} is obtained for the kdistribution having min(k_{max}  k_{min}). The corollary implies that the kdistribution has to be the most 'flat'like one as compared to other distributions resulting in other Sum_{TOTAL} values. Then M_{Σ1/π(p)} = 2.813839146... > 2.813838890... for min(k_{max}  k_{min}) = 5.047135130... on the basis of the 51 known Mersenne prime exponents. Note: Similar results using the same approach can be obtained also for the other Mersenne constants, like M_{Σ1/p}, etc. M_{Σ1/π(p)} was chosen as an illustration because the Wynn's Epsilon Method has a better convergence for it as compared to M_{Σ1/p}. (* Wolfram code *) Mexponent = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933, 0}; Mmax = 51; SumArray = ConstantArray[0, Mmax]; k = ConstantArray[0, Mmax]; sum = 0; m = 1; While[m <= Mmax, sum = sum + 1/PrimePi[Mexponent[[m]]]; SumArray[[m]] = sum; m++]; limitmin = SumArray[[Mmax]]; limit = limitmin; limitlbest = limitmin; kbest = 100; nbest = 0; n = 0; While[n <= 10000, k[[1]] = limit; m = 2; While[m <= Mmax, k[[m]] = PrimePi[Mexponent[[m]]]*(limit  SumArray[[m  1]])  1; m++]; kmin = Min[k[[1 ;; Mmax]]]; kmax = Max[k[[1 ;; Mmax]]]; If[kmax  kmin < kbest, limitbest = limit; kbest = kmax  kmin; nbest = n;]; limit = limitmin + 0.0000000001*n; n++]; Print[SetPrecision[N[limitbest], 10], ", ", SetPrecision[N[kbest], 10], ", ", nbest]; 

20210822, 21:12  #33 
May 2018
2·97 Posts 
Assuming that M_{Σ1/π(p)} = 2.813839146..., one could roughly estimate that 154,900,853 <= M_{52} <= 540,044,243 for k_{min} = 1.236136005... and k_{max} = 6.283271135...
Even if k_{min} = 1, then 137,615,273 <= M_{52}. #, M_{exponent}, π(M_{exponent}), k 52, 154900853, 8704305, 1.236136005 < k_{min} 52, 540044243, 28350608, 6.283271135 < k_{max} (* Wolfram code *) limit = 2.81383914655656131031946642906; Mexponent = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933, 0}; Mmax = 51; SumArray = ConstantArray[0, Mmax + 1]; k = ConstantArray[0, Mmax + 1]; sum = 0; m = 1; While[m <= Mmax, sum = sum + 1/PrimePi[Mexponent[[m]]]; SumArray[[m]] = sum; m++]; k[[1]] = limit; m = 2; While[m <= Mmax, k[[m]] = PrimePi[Mexponent[[m]]]*(limit  SumArray[[m  1]])  1; m++]; kmin = Min[k[[1 ;; Mmax]]]; kmax = Max[k[[1 ;; Mmax]]]; m = Mmax; If[(limit  SumArray[[m]]) > 0, k[[m + 1]] = kmin; Mexponent[[m + 1]] = Prime[Round[(1 + k[[m + 1]])/(limit  SumArray[[m]])]]; SumArray[[m + 1]] = SumArray[[m]] + 1/PrimePi[Mexponent[[m + 1]]]; Print[m + 1, ", ", Mexponent[[m + 1]], ", ", PrimePi[Mexponent[[m + 1]]], ", ", SetPrecision[N[k[[m + 1]]], 10], " < kmin"]; k[[m + 1]] = kmax; Mexponent[[m + 1]] = Prime[Round[(1 + k[[m + 1]])/(limit  SumArray[[m]])]]; SumArray[[m + 1]] = SumArray[[m]] + 1/PrimePi[Mexponent[[m + 1]]]; Print[m + 1, ", ", Mexponent[[m + 1]], ", ", PrimePi[Mexponent[[m + 1]]], ", ", SetPrecision[N[k[[m + 1]]], 10], " < kmax"];]; 
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