Sorry, Ed. I discovered a couple of typos in the message I sent. Please ignore that message. Here's the corrected message. That is a cute result. It seems clear that there should be some u_0 where product(n=1...N, (u + sin n)) ->0 if uu_0, but who would have thought that u_0 would be rational! We have product(n=1...N, (5/4 + sin n)) = exp sum(n=1...N, ln (5/4 + sin n)) The common difference of the value of n, viz 1, is incommensurate with the periodicity of the summand, viz 2 pi. Thus, the law of large numbers applies. Thus, for any -1<=y_0oo, the proportion of values of n in the partial sum sum(n=1...N ln (5/4 + sin n)) where y_0oo [sum(n=1...N, ln (5/4 + sin n))] / N = int(x=-pi/2...pi/2, ln (5/4 + sin x) dx) / pi ___ [1] We have the following indefinite integral equation: d x 2 p tan (ax/2) + q int(------------) = ------------- arctan ---------------- (if p>q) p + q sin ax a sqrt(p²-q²) sqrt(p²-q²) therefore, setting p=u, q=1, a=1, d x 2 u tan (x/2) + 1 int(---------) = ---------- arctan --------------- ___ [2] u + sin x sqrt(u²-1) sqrt(u²-1) Define, for u>1, I(u) = int(x=-pi/2...pi/2, ln (u + sin x) dx) Differentiating under the integral sign, d I(u) dx ------ = int(x=-pi/2...pi/2, --------- ) d u u + sin x [ 2 u tan (x/2) + 1 ]x=pi/2 = [ ---------- arctan --------------- ] from eqn [2] [ sqrt(u²-1) sqrt(u²-1) ]x=-pi/2 2 [ u tan (x/2) + 1 ]x=pi/2 = ---------- [ arctan --------------- ] sqrt(u²-1) [ sqrt(u²-1) ]x=-pi/2 2 [ u+1 -u+1 ] = ---------- [ arctan ---------- - arctan ---------- ] ___ [3] sqrt(u²-1) [ sqrt(u²-1) sqrt(u²-1) ] Now, u+1 -u+1 1-u² ---------- * ---------- = ---- = -1 sqrt(u²-1) sqrt(u²-1) u²-1 and if tan x_0 = y_0, tan x_1 = y_1, and y_0 y_1 = -1, then x_0 and x_1 differ by pi/2. Thus, from equation 3, d I(u) pi ------ = ---------- d u sqrt(u²-1) Integrate with respect to u. I(u) = pi ln(u+sqrt(u²-1)) + c ___ [4] where c is the constant of integration. Going back to the definition of I, I(u) = int(x=-pi/2...pi/2, ln (u + sin x) dx) = int(x=-pi/2...pi/2, ln (u (1 + (sin x)/u)) dx) = int(x=-pi/2...pi/2, ln u + ln (1 + (sin x)/u) dx) = pi ln u + int(x=-pi/2...pi/2, ln (1 + (sin x)/u) dx) ___ [5] As u->oo, (sin x)/u -> 0 => 1 + (sin x)/u -> 1 => ln (1 + (sin x)/u) -> 0 convergence being uniform in all cases, thus the integral in equation 5 approaches 0. Thus, putting equations 4 and 5 together, as u->oo pi ln u -> pi ln(u+sqrt(u²-1)) + c => c = lim_u->oo [pi ln u - pi ln(u+sqrt(u²-1))] = pi lim_u->oo [ln u - ln(u+sqrt(u²-1))] = pi lim_u->oo [ln u - ln(u+sqrt(u²))] = pi lim_u->oo [ln u - ln(2 u)] = -pi ln 2. Therefore I(u) = pi ln(u+sqrt(u²-1)) - pi ln 2 In particular, I(5/4) = pi ln[5/4+sqrt(25/16-1)] - pi ln 2 = pi ln[5/4+sqrt(9/16)] - pi ln 2 = pi ln(5/4+3/4) - pi ln 2 = 0. Thus, going back to equation 1, lim_N->oo [sum(n=1...N, ln (5/4 + sin n))] / N = 0 => sum(n=1...N, ln (5/4 + sin n)) = o(N) As stated above, the law of large numbers applies. Moreover, ln (5/4 + sin n) is bounded by ln 1/4 and ln 9/4, therefore, any short-term effects whereby large terms dominate the sum are always smoothed out in the long run as N->oo. Thus sum(n=1...N, ln (5/4 + sin n)) = O(1) and sum(n=1...oo, ln (5/4 + sin n)) does not diverge. Thus product(n=1...oo, (5/4 + sin n)) doesn't diverge or have 0 as a limit. Moreover, if there were a non-zero limit L such that lim_N->oo product(n=1...N, (5/4 + sin n)) = L then lim_N->oo sum(n=1...N, ln (5/4 + sin n)) = ln L => lim_N->oo sum(n=N...oo, ln (5/4 + sin n)) = 0 => lim_n->oo ln (5/4 + sin n) = 0 => lim_n->oo 5/4 + sin n = 1 which is false. Thus product(n=1...N, (5/4 + sin n)) has no non-zero limit as N->oo. To summarise, product(n=1...oo, (5/4 + sin n)) does not diverge or have a limit. -- Richard Sabey ---------------------------------------------------------------------- Dear Mr Pegg Here is a short TeX-file with a few remarks concerning Tommy's problem. \hsize=12cm \parindent=0pt The natural numbers are equidistributed mod $2\pi$. This implies that for any continuous $2\pi$-periodic function $f$ we have $$\lim_{N\to\infty}{1\over N}\sum_{n=1}^N f(n) ={1\over 2\pi} \int_0^{2\pi} f(x)\,dx\ .$$ In particular, $$\lim_{N\to\infty}{1\over N}\sum_{n=1}^N \log\Bigl({5\over 4} + \sin(n)\Bigr) = {1\over 2\pi}\int_0^{2\pi}\log\Bigl({5\over4} +\sin(x)\Bigr)\,dx =:A\ .$$ The integral $A$ seems to be nonelementary, but anyway, it computes to 0. (This has to do with the Pythagorean (3,4,5)-triangle!) It follows that in order to solve the original problem one would have to use finer results about equidistribution. \end Sincerely yours Christian Blatter --------------------------------------- Ed, I’ll be interested to see a proof of that, since it appears for all intents and purposes to go to infinity. To be convergent, one would think the increments of the final value as n increases would diminish. But this equation has a repeating pattern of geometric increments and decrements, of which the majority are increments. It would seem to me that the equation steadily increases geometrically, thus approaching infinity. That’s my initial take on the problem, but I never did any math past Calculus 3 in college so I’m probably missing something. Looking forward to the next update, Clint Weaver --------------------------------------- Tommy wants proof that the product, for n=1 to infinity, of (5/4 + sin(n)) goes to neither zero nor infinity. I can offer a start. If we take the logarithm of this expression, we need that the logarithm goes to neither -infinity nor +infinity, but instead stays finite. The log becomes the sum, for n=1 to infinity, of ln(5/4 + sin(n)). Now this clearly never converges to a value, but we can find the mean value of the terms in the sum by collapsing all the points into the single sine wave from -pi to +pi, and considering that, due to the irrationality of pi, sin(n) over the integers hits all parts of this sine curve equally often. In other words, the mean value is the integral from x = -pi to +pi of ln(5/4 + sin(x)), divided by the 2pi length of the interval. Now I don't know how to integrate this (maybe fool around with integration by parts, where in the first step dv = dx and you get a still messy integral involving only x and sin(x) and cos(x), or just let Mathematica grind away on it for a while). For the proof to work, we need the mean value of the logarithmic terms to be 0. If it is non-zero, then the sum will diverge, and the infinite product will correspondingly head to zero or infinity. Joseph DeVincentis --------------------------------------- the integral i mentioned earlier today exp (int ( sin -5/4)) is indeed -1 . also interesting is the product taken over the primes , lineairy variants etc eg sin(p)+5/4 , sin(2n)+5/4 (behaves mainly the same ) sin ((f(n)) -(f ' (n))^-a +5/4) especiallly sin(mn)+5/4. that sin(355n) +5/4 diverges is logical . 355 is a approximate denomitor ( if im correct srr for my english ) for a rational approcing pi . most cases where sec (m ) is high, sin(mn) +5/4 diverges. but why does m =142 diverges and 143 does not !! ? also sin (n+m) +5/4 should behave the same for all positive real m according to the integral ( integral is taken over period so ) if m is rational of course. Secondly id like to wunder if that "integral estimate " only works for products with a flat upper bound and lower bound with attrackts it strongly ( for the product itself and the function which is " producted " (mulitplied ? ) , I fear that at first sight. Have you seen A098066 yet ? also sec(n+1) is never bigger than (n+1) (n+1) i believe strongly. I even expect a connection with sin+5/4. greetz Tommy >An item to look at: > >http://www.math.niu.edu/~rusin/known-math/99/contour_int > >--Ed Pegg Jr ------------------------------------------------------------------