Let H(n) denote the sum of the first n terms of the harmonic series, i.e., 1 1 1 1 H(n) = 1 + --- + --- + --- + ... + --- 2 3 4 n It's well known that H(n) -> log(n) + gamma as n -> inf where gamma = 0.57721... is Euler's constant and log is the natural logarithm. Interestingly, it appears that the discrepancy between H(n) and log(n)+gamma can be computed very directly from the Bernoulli numbers. This relation not only allows us to determine H(n) very precisely, it can also be used to compute the value of gamma from a known value of H(n). For any integer n>1 it appears that inf B_k H(n) = log(n) + gamma - SUM ----- (1) k=1 k n^k where B_k denotes the kth Bernoulli number, i.e., B_0 = 1 B_1 = -1/2 B_2 = 1/6 B_3 = 0 B_4 = -1/30 and so on. The summation _appears_ to rapidly converge, so that just ten non-zero terms are needed to give H(256) = 6.12434496281728... Conversely, given (for example) the fact that H(8) = 2283/840, we can easily compute 2283 inf B_k gamma = ---- - 3 log(2) + SUM ----- (2) 840 k=1 k n^k Again taking just the first ten non-zero terms of the summation, this formula gives gamma = 0.577215664901533.... which is correct to 15 decimal places. Of course, the value of log(2) is easily computed using the formula / 1 + 1/3 \ / 1 1 1 \ log(2) = log( --------- ) = 2 ( --- + ----- + ----- + ... ) \ 1 - 1/3 / \ 1*3 3*3^3 5*3^5 / Anyway, more digits of gamma can be determined simply by using a larger value of n (e.g., n=16) with the same number of terms in the Bernoulli summation. However... When I said the summation _appears_ to converge very rapidly, I was referring to the fact that the terms initially become extremely small as k increases. This is because the denominator k n^k increases rapidly whereas the numerator B_k actually gets smaller...for a while. But in the long run the values of the Bernoulli numbers increase at a super-exponential rate, as can be seen from the inequality 2(2k)! | B_(2k) | > ----------- (3) (2 PI)^(2k) Therefore, in the long run the numerators of the terms in the summation rise faster than the denominators, so the summations in equations (1) and (2) actually diverge. This strikes me as a very interesting example of asymptotic divergent series. The reader may have noticed that (1) is given directly by the Euler- Maclaurin summation formula. I didn't initially recognize this as an EM summation because of the round-about path by which I arrived at it. For any positive integers b and N you can construct a Nth order linear recurring sequence approximating the values of H(b^k), k=0,1,2,... For example, setting h(k) = H(2^k) we have the (approximate) 3rd order recurrence 2h(k) - 5h(k-1) + 4h(k-2) - h(k-3) = 0 Taking the exact initial values h(0)=1, h(1)=3/2, and h(2)=25/12, the closed-form solution of this recurrence gives the (approximate) values 2 2 1 h(k) = --- k + --- + ----- 3 3 3*2^k Of course, this is exact only for the three initial values (k=0,1,2), so let's call this function h3(k). Similarly the fourth order "harmonic recurrence" gives the function 218 110 17 64 h4(k) = --- k + --- + ------ - ---------- 315 189 35*2^k 945*2^(2k) and so on. Obviously the first j values of hj(k) are the exact values h(k) of the harmonic series. Going to higher and higher order formulas it quickly becomes apparent that the coefficient of k is approaching ln(2), and the constant coefficient is approaching gamma. For example, the constant coefficient of h7(k) is 330229884852951965355239887485346 --------------------------------- = 0.5772156... 572108310987828197293898319807675 Also, the coefficients of the subsequent terms approach B_n/n, so it would be natural to jump to the conclusion of equation (1)...until realizing that the summation diverges.

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