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I am curious about the size of the sum of the first Graham's number terms of the harmonic series. The harmonic series is a well-known mathematical series, and Graham's number is an incredibly large number. So, I wonder what would happen if we summed the first Graham's number terms of the harmonic series. I find the idea of calculating this sum to be aesthetically pleasing, as the harmonic series is a simple and elegant series, and Graham's number is a kind of mathematical monster. I am also motivated by the challenge of calculating such a large sum, as it would require a deep understanding of mathematics and computing. I believe that it would be a good way to measure the size of Graham's number with respect to the size of the harmonic series sum.

Mahmoud albahar
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    The sum is approximately the natural logarithm of Graham's number , but this number is so huge that taking the logarithm has virtually no effect. The sum is basically Graham's number itself , but it is not an integer (this can be proven for the sum of the first $n$ terms for every integer $n>1$). – Peter Sep 04 '23 at 10:36
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    Letting ^^ denote tetration (operation that is repeated exponentiation), we essentially have the sum of 10 ^^ n many harmonic terms equal to 10 ^^ (n-1) (and 10 can essentially be replaced by any other positive integer greater than 1 and less than a million). So for values of $n$ such as $n=10^{10^{10}}$ (which is only 10 ^^3, by the way), the sum of 10 ^^ n many harmonic terms is essentially equal to 10 ^^n (because $10^{10^{10}} - 1$ is essentially equal to $10^{10^{10}}).$ For the repeated tetration operation, the form of the number would be essentially unchanged. (continued) – Dave L. Renfro Sep 04 '23 at 10:40
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    So WAY before we get to Graham's number, looking at the sum of a certain number of terms of the harmonic series essentially doesn't change the number in any reasonable notational sense. You might find some sci.math essays I wrote back in 2002 of interest, the links of which are in this mathoverflow answer, especially Section D of the one titled "BIG NUMBERS #3". – Dave L. Renfro Sep 04 '23 at 10:42
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    The Graham-sequence grows incredibly quick , it is defined as $G_0=4$ , $G_{n+1}=3\uparrow ^{G_n} 3$ for $n\ge 0$. Already $G_1=3\uparrow^4 3$ is incomprehensible in size , so Dave's comment applies already for such a number. Graham's number is $G_{64}$ unimaginably larger then $G_1$ which is already a monster. – Peter Sep 04 '23 at 10:45
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    Concerning the harmonic series , it grows very slow , but at least with $\ln(n)$ , the sum of the reciprocals of the primes grows muuch slower , with $\ln(\ln(n))$ , but it is still divergent. – Peter Sep 04 '23 at 10:49
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    You might also see this question – Ross Millikan Sep 04 '23 at 13:14
  • @Peter: It is not known whether there are an infinite number of even perfect numbers. However, if there were, and we calculated the sum of the first Graham's number of reciprocals of even perfect numbers, then it could be handled. I hope there will be an answer to my comment after one or two centuries. – Mahmoud albahar Sep 04 '23 at 19:21
  • @Peter: I said after one or two centuries because these are complicated things in mathematics. – Mahmoud albahar Sep 04 '23 at 19:22
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    Note that the sum of the reciprocals of a sequence does not always diverge. For the powers of $2$ , we have $1/2+1/4+1/8+\cdots=1$ and for the perfect even numbers , even if there are infinite many , we will have a very small sum as well. To find out whether there are infinite many even perfect numbers is probably out of reach , but we can surely bound the sum of their reciprocals. – Peter Sep 05 '23 at 07:48

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