• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 sum(e - eta^^k): convergence or divergence? Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 07/20/2010, 10:30 AM (This post was last modified: 07/20/2010, 10:33 AM by Gottfried.) From some fiddlings with the slog-subject I came across the question, whether this is divergent or convergent: using $\eta = e^{1/e} = 1.44466786...$ the sum: $S = \sum_{k=0}^{\infty} ( e- \eta\^ \^^k )$ Clearly the sequence of terms tends to zero because e is the fixpoint of iteration and in a first guess I thought that also the series converges. But the convergence of the sequence is slow and one needs a lot of terms to see a promising trend. What I did is to look at the sequence of partial sums, each from zero to 2^n, $s_n = \sum_{k=0}^{2^n} ( e - \eta\^ \^^k )$ and that sequence {s_n} seem to increase, even slightly more than linear. at least if I look at the partial sums up to n = 12. Here are the partial sums and the differences of order 1 to 3: Code:.  n  s_n          d1_n=s_n - s_(n-1)   d2_n=d1_n-d1_(n-1)   d3_n .  1  4.00875692871  4.00875692871     4.00875692871      4.00875692871 .  2  5.58578587004  1.57702894134    -2.43172798737     -6.44048491607 .  3  7.77673131247  2.19094544242    0.613916501085      3.04564448845 .  4  10.5161613469  2.73943003447    0.548484592045   -0.0654319090397 .  5  13.6651189223  3.14895757539    0.409527540923    -0.138957051122 .  6  17.0811305635  3.41601164122    0.267054065824    -0.142473475099 .  7  20.6561843799  3.57505381638    0.159042175165    -0.108011890659 .  8  24.3207908875  3.66460650761   0.0895526912324   -0.0694894839327 .  9  28.0341817806  3.71339089304   0.0487843854235   -0.0407683058090 . 10  31.7736430757  3.73946129511   0.0260704020760   -0.0227139833474 . 11  35.5268806871  3.75323761140   0.0137763162852   -0.0122940857909 . 12  39.2873487126  3.76046802550  0.00723041410300  -0.00654590218217 How could we prove the divergence/convergence of the series S? Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 Joined: Feb 2009 07/20/2010, 09:24 PM (This post was last modified: 07/20/2010, 10:19 PM by tommy1729.) reminds me a bit of paris constant. maybe this is an analogue. also , it reminds me of your " tiny limit curiosity " , it seems logical to me that if the equation doesnt hold for base eta , then the sum should diverge. but powers of 1 remain 1 ... thus probably it diverges. regards tommy1729 bo198214 Administrator Posts: 1,394 Threads: 91 Joined: Aug 2007 07/24/2010, 11:38 PM (07/20/2010, 10:30 AM)Gottfried Wrote: using $\eta = e^{1/e} = 1.44466786...$ the sum: $S = \sum_{k=0}^{\infty} ( e- \eta\^ \^^k )$ Clearly the sequence of terms tends to zero because e is the fixpoint of iteration and in a first guess I thought that also the series converges. But the convergence of the sequence is slow and one needs a lot of terms to see a promising trend. Perhaps you can post the open problem in the open problems survery. I dont think it is too difficult, however didnt find a proper proof yet. tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 Joined: Feb 2009 08/16/2010, 11:05 PM i almost forgot about this , but when considering paris constant i think i stumbled upon a proof. the assumed proof shows divergence ( as i first suspected ). first i show that it cannot converge at an exponential rate. this is done by noting the koenigs analogue : note that lim k -> oo eta^^k ^ z is in the neigbourhood of eta^^(k-1) and eta^^(k+1) if z is in the neighbourhood of eta. hence lim k -> oo eta^^k ^ z ~ eta ^^k koenigs then becomes lim k-> oo (eta^^k - e) / Q^k where Q is the derivate of eta^x at x = e ( since e is fixpoint ). D eta^x = eta^x / e => eta^e / e = e/e = 1. thus we arrive at Q = 1 and lim k-> oo (eta^^k - e) / 1 hence eta^^k - e shrinks slower than exponential. ( for bases > eta , Q > 1 and for bases < eta , Q < 1 hence this can be used to prove the div or conv for other bases easily. ) since eta^^k - e shrinks slowly we can approximate it well with polynomials and thus we construct a taylor series. r_k = -(eta^^k - e) = e - eta^^k a strictly negative term sum converges iff its corresponding positive term sum converges and vice versa where corresponding means all terms multiplied by -1. r_k = e - eta^^k r_k+1 = e - eta^^k+1 = e - eta^eta^^k r_k+1 = e - eta^(-r_k + e) ( because eta^^k = -r_k +e ) r_k+1 = e - eta^(-r_k + e) = e - e eta^-r_k ( because eta^e = e ) r_k+1 = e (1 - eta^-r_k) and this is a pretty selfref about how fast r_k grows to e , now use taylor : r_k+1 = e ( r_k/e - r_k^2 /(2e^2) + r_k^3/(6e^3) + O(r_k^4) ) now §r_k+1§ - r_k = r_k^2 /(2e) + r_k^3/(6e^2) + O(r_k^4) compare §e/(r+1)§ - e/r = e^2/(2e r^2) => 1/(r+1) - 1/r = - r^-2 / 2 => 1/r - 1/(r+1) = r^-2 / 2. which implies that since eta^^k - e shrinks slower than exponential , r_k seems close to 2e/k since by the above we can remove the /2 part by doubling our estimate of e/r : 2/r - 2/(r+1) = r^-2 2/2 = r^-2. and of course 1/r - 1/(r+1) = r^2 approximately for 1/r hence r_k seems between 2e/(sqrt(r^2 + 3r)) + C/r^3 and 2e/(sqrt(r^2 -3r)) + C/r^3. thus r_1 + r_2 + r_3 + ... diverges within the order of O(log(x) + C/x^2). Q.E.D. tommy1729 Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 08/17/2010, 11:45 AM (This post was last modified: 08/17/2010, 11:52 AM by Gottfried.) In the newsgroup news://sci.math I got a nice and concise answer by Prof. Israel. At 16.08.2010 with the subject: "Series : divergent or convergent?" I gave t=e=exp(1), b=t^(1/t) and the notation for the residual-term r_k = t - b^^k as example. This is the answer: (I inserted the correction of a wrong sign) Code:r_{k+1} = t - b^(b^^k) = t - b^(t - r_k)         = t - t b^(-r_k)  (since t = b^t)         = t (r_k ln(b) + O(r_k^2))   By the ratio test, the series will converge if |t ln(b)| < 1.   Since b^t = t says t ln(b) = ln(t), this is equivalent to 1/e < t < e. Your case t=e is on the boundary of this, so we need another term. r_{k+1} = t (r_k ln(b) - r_k^2 ln(b)^2/2 + O(r_k^3))         = r_k - r_k^2/(2 e) + O(r_k^3)              This fits with r_k ~ 2e/k, which would indicate that the sum diverges. [second post] In fact, I believe we should have      r_k ~= 2e/(k + ln(k)/(3 - 1/k))  as k -> infty. Again, it diverges. -- Robert Israel     Department of Mathematics     University of British Columbia Vancouver, BC, Canada I think that solves the problem. @Henryk: Shall I still copy the problem into the TPID-section? (or the math-facts/does this still exist?) Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 Joined: Feb 2009 08/17/2010, 10:33 PM didnt read my posts yesterday ? i probably proved 2 of your statements with basicly the same method. i dont know where 2e/(k + ln(k)/(3 - 1/k)) is coming from btw and i dont see it explained. it seems robert didnt show that e-r_k doesnt converge at exp speed and hence he merely did a taylor series recursion , though i admit so did i finally. furthermore sum 1/(k*log^3(k)) converges so the proof certainly needs to be made more rigorous. i used the koenigs analogue to prevent wild solutions to the taylor series recursion , however a strong proof or construction of a solution to the taylor series recursion to prevent e.g. r_k ~~ 1/(k*log^3(k)). is needed. robert merely gave a recursion , i did a bit more. but perhaps not enough. im thinking about improving what i wrote yesterday. maybe replace koenigs analogue with a better formula analogue. and partially replacing taylor ( after r_3 term ) with something better. and robert is not a full prof i believe. dont get me wrong , i do not wish to belittle robert , i respect him and supported him in the past. but i dont like you skipping my reply ... and ignoring my other potential proof. forgive my anger , but i am a man of honor. maybe you meant to reply at my posts later ... do me a favor to make up for it and read my other potential proof in ' tiny limit curiosity '. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 Joined: Feb 2009 08/17/2010, 11:05 PM and i now read the thread on sci.math and it seems your schroeder function idea is ' borrowed ' from my koenigs analogue posted yesterday here ... « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Improving convergence of Andrew's slog jaydfox 19 40,448 07/02/2010, 06:59 AM Last Post: bo198214 Convergence of matrix solution for base e jaydfox 6 13,933 12/18/2007, 12:14 AM Last Post: jaydfox

Users browsing this thread: 2 Guest(s)