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+--- Thread: sum(e - eta^^k): convergence or divergence? (/showthread.php?tid=476)
sum(e - eta^^k): convergence or divergence? - Gottfried - 07/20/2010
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 -0.00654590218217How could we prove the divergence/convergence of the series S?
Gottfried
RE: sum(e - eta^^k): convergence or divergence? - tommy1729 - 07/20/2010
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
RE: sum(e - eta^^k): convergence or divergence? - bo198214 - 07/24/2010
(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.
RE: sum(e - eta^^k): convergence or divergence? - tommy1729 - 08/16/2010
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
RE: sum(e - eta^^k): convergence or divergence? - Gottfried - 08/17/2010
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, CanadaI 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
RE: sum(e - eta^^k): convergence or divergence? - tommy1729 - 08/17/2010
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
RE: sum(e - eta^^k): convergence or divergence? - tommy1729 - 08/17/2010
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 ...