From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: A help with the limit of a series, please Date: 22 Dec 1998 20:01:19 -0500 In article <75p8f4$516$1@nnrp1.dejanews.com>, wrote: [...] >> > Can anyone help me to determine the limit of the series [edited] sum[n=2 to infinity] 1/(n * (ln(n))^2) >> > I could prove it's convergent, but couldn't determine it's limit >analytically. >> > By ln (n), I mean the natural logarithm of n. [...] >Using an Excel spreadsheet, I added the terms from 2 to 2000 and got >1,978183803. This series converges very slowly, and illustrated the advantages of preparing a problem using some theory before crunching the numbers. Using simple tricks related to Integral Test, we can improve the convergence of the sequence of partial sums b(n) = 1/(2*ln(2)^2) + ... + 1/(n*ln(n)^2) by considering the trapezoidal corrections s(n) = b(n) + 1/ln(n) - 1/(2*n*ln(n)^2) and midpoint corrections t(n) = b(n) + 1/ln(n+1/2) Since the corrections converge to 0 as n grows, the sequences {s(n)} and {t(n)} converge to the limit of {b(n)}. Theory will reveal that s(n) will increase and t(n) will decrease, thus giving us a clear idea about accuracy. Here is an edited printout form MATLAB: n s(n) t(n) 200 2.10974269900978 2.10974285204124 400 2.10974278188489 2.10974281088421 600 2.10974279381146 2.10974280494235 800 2.10974279745107 2.10974280312705 1000 2.10974279898485 2.10974280236161 2000 2.10974280078141 2.10974280146450 Since I even arranged the summation from the smallest to the largest term, to reduce round-off errors, you can be sure that the sum is approximately 2.109742801, correctly rounded. More calculations will tell you that to obtain b(n) > 2 (quite a modest requirement), you will need n >= 9065. Worse yet, to obtain b(n) > 2.1, you would be facing n > 3.7 * 10^44 (you wouldn't live that long). (The remainder is of order of magnitude 1/ln(n), decreasing very slowly.) More systematic treatment of convergence acceleration of series of types similar to this one is known as Euler-Maclaurin Formula. Cheers, ZVK(Slavek).