Since lnx is strictly increasing and n+1>1, then ln(n+1)>ln1=0, thus xn>0.
Further, since lnx tends to infinity as x tends to infinity, then ln(n+1) tends to infinity as n tends to infinity, i.e. xn diverges to infinity.
Also, xn+1-xn= ln(n+2)- ln(n+1)=ln((n+2)\(n+1)), and since (n+2)\(n+1) tends to 1 as n tends to infinity, then ln((n+2)\(n+1)) tends to ln1=0, and thus xn+1-xn tends to 0 as n tends to infinity.
Further, let (yn) be a strictly increasing sequence of natural numbers.
Then xyn is a subsequence of xn, and xyn-xn= ln(yn+1)- ln(n+1)=ln((yn+1)\(n+1)).
Choosing yn=mn, where m is a natural number at least 3, we have that (yn) is a strictly increasing sequence of natural numbers and xyn-xn= ln((mn+1)\(n+1)).
As n tends to infinity, the sequence (mn+1)\(n+1) tends to m, and thus xyn-xn tends to lnm, which is greater than 1, as m is at least 3.
To include 1 in the limits of xyn-xn, we may divide the initial sequence by ln3, i.e.
we may consider the sequence xn=ln(n+1)\ln3.
The new sequence has strictly positive terms, diverges to infinity, it still has the property that xn+1-xn tends to 0, and if yn=mn, then xyn-xn tends to lnm\ln3, which is 1 if m=3, and >1 if m>3.
Yes, dear Issam Kaddoura , but you have given just an example. I think that George's problem is to prove the existence of the subsequence k(n) for every c >0 separately, whatever is the sequence xn satisfying the assumption
(*) . . . . 0 < xn ↗ ∞ such that xn+1 - xn → 0 as n → ∞ .
My suggestion is to use the following structure:
THEOREM. Under (*), for every c ≥ 0, there exists a sequence k(n) such that
(**) . . . . . xk(n) - xn → c .
Proof: The claim holds for c=0 trivially. For c>0 one can start with
LEMMA. The claim holds for c=1.
(I am omitting the proof of the Lemma since it's claim is contained in the fact formulated by George).
Now, for any other c>0, let us introduce yn := xn/c. This sequence satisfies the assumption of the theorem too. By the lemma for some sequence k(n) we have
(***) . . . . yk(n) - yn → 1
Multiplying statement (***) by c we arrive at (**).
Thank you all for very good contributions; and also thank you for noticing the typo in my answer, I just corrected it.
P.S. I still have a small issue with my own solution, namely: is k(n) well defined, i.e., is k(n) always finite, or it depends upon c smaller/larger than 1? Please take a look...
You need a little more. To get a subsequence you must have k(n) strictly increasing. That does not follow from the construction. You need to find some way to make sure you have k(n) < k (n + 1).
It is bounded above, and in fact finite, because for a fixed n, x_k - x_n has limit infinity (as k goes to infinity) and after a finite number of values will be > c. So max can be used.
Dear George Stoica , we can call the sequence {xk(n)} a stuttering subsequence * of {xn}. This would give birth to a new notion (unless it already exists). :)
* it is sure that {xk(n)} is non-decreasing because {xn} is increasing
The set { k ≥ n : x_k - x_n < c } may be empty up to some order n_0. Therefore we have to start from a given n_0 so that $x_{k+1} - x_k < c$ given by the convergence.
I think that we can make the $x_{k(n)}$ a real subsequence, as follows:
Choose $k_0$ so that $x_{k_0 + 1} - x_{k_0} < c$. Then take $k_1 = \sup{ k ≥ k_0 : x_k - x_{k_0} < c }$. Now by induction take $k_{n+1} = \sup{ k ≥ k_n : x_k - x_{k_n} < c }$. The sequence $(k_n)$ is increasing, since $k_{n+1}$ is at least $k_n + 1$. Now continu at noticed Stoica.
just for having all our remarks in one, in the formulation I would replace the problem by a request of a proof of the following
THEOREM: Let 0 < xn ↗ ∞ be such that
(*) . . . . . . . xn+1 - xn → 0 as n → ∞ .
Then, for every c> 0 there exists a sequence k(n) such that xk(n) - xn → c as n → ∞ . The sequence can be chosen as nondecreasing.
PROOF (by construction): Let us fix c>0. By the assumed monotonicity and unboundedness of sequence xn , for every n there exists k ≥ n such that xk > xn + c. Such k is greater than n. Thus let us define k(n) ≔ min{ k≥n : xk > xn + c } [ which is greater than n and is equal to max{ k ≥ n : xk ≤ xn + c } +1 ]
Obviously, k(n) is nondecreasing. Moreover we have:
limn →∞ ( xk(n) - xn ) = c.
Indeed, otherwise there would exist an increasing sequence n(ℓ)→∞ , as ℓ→∞ , and such that for some positive ε >0
(**) . . . . . . . xk(n(ℓ)) - xn(ℓ) ≥ c + ε for ALL ℓ
On the other hand, by the definition of k(n), for every n we have xk(n)-1 - xn ≤ c for ALL n , in particular
(***) . . . . . . . xk(n(ℓ))-1 - xn(ℓ) ≤ c for ALL ℓ
Subtracting the inequalities (**) and (***) side by side, we arrive at
Thank you, dear George, for the recommendation, but it is due to all P.T. Contributors of your intersting question.
I have just collected and exploited the main issues, remarks, suggestions, subquestions etc. Even those seemingly/allegedly wrong statements help those who care in reaching the rightest true:-)
There is already a name for it: subnet. Sequences are nets and as such have subnets. some of which may not be subsequences. When they are not it is exactly because of repeating terms. But your name of the notion is more revealing!
The assertion is true for the famous harmonic series Hn for all real values c ≥ 0. Namely,
Hn: = 1+1/2+1/3+ + +1/n (n = 1,2,…).
It is well known that
Hn ≈ ln n + γ + 1/(2n) as n → ∞ (1),
where γ ≈ 0.577215… is the Euler–Mascheroni constant. For any nonnegative real numbers a, put k(n) = n + [an], where [x] denotes the greatest integer not exceeding x. Then from (1) we find that
Hk(n) - Hn = Hn+[an] - Hn ≈ ln(n + [an]) - ln n - [an]/(2n(n + [an])) = ln(1+ [an]/n) - [an]/(2n(n + [an])) as n → ∞,
and hence since an ≤ [an] < an +1, we have
Hk(n) - Hn → ln(1+a) as n → ∞.
Finally, taking a = ec -1 with c ≥ 0 in the above limit relation, we obtain
>> If x_(n + 1) - x_n > c then the only possible value for k(n) is n. You said that above. If we want x_k(n) to be a subsequence of x_n, once we have k(n) = n for some n then k(m) = m for all m
Let a be a fixed real number such that ) 0 < a < 1, and let xn ≈ na, i.e., lim_(n → ∞) (xn/na) = 1. Let c ≥ 0 be any nonnegative real number. Put k(n) = n + [cn1-a/a] (n = 1,2,… ), where [x] denotes the greatest integer not exceeding x. Then it can be proved that
xk(n) - xn → c as n → ∞.
For example, if xn ≈ √n, then k(n) = n + [2c√n], then it is easy to show that
xk(n) - xn = (n + [2c√n])1/2 - n1/2 → c for all c ≥ 0.
Hence, the proposed assertion is true for all sequencec (xn) with xn ≈ na (0 < a < 1) with respect to all real values c ≥ 0.
Numeruos computations in Wolfram’s Mathematica 9 sugges the following extensions of results attributed in my previous two answers, given as follows.
Let a and b be fixed real numbers such that 0 < a ≤ 1, and let xn ≈ na/lnb n, i.e., lim_(n → ∞) (xn lnb n/na) = 1. Let c ≥ 0 be any nonnegative real number. Put
k(n) = n + [cn1-aln n/a] (n = 1,2,… ), (1)
where [x] denotes the greatest integer not exceeding x. Then
xK(n) - xn → c as n → ∞.
Remark 1. Notice that the values of the sequence k(n) given by (1) does not depend on b.
Remark 2. The case when a = 0, b = -1, i.e., the sequence xn ≈ ln n is not included here, but this case is in fact solved in my first answer concerning the harmonic series Hn:= 1 +1/2 + 1/3 + +1/n ≈ ln n with related required subsequence k(n) = n + [(ec -1)n] (c ≥ 0).
Remark 3. The cases when a = 0 and (then must be) b < 0 also are not included here, i.e., the cases of the logarithmic sequences of the form xn ≈ ln-b n with any real value b < 0. A computation in Mathematica 9 (which also can be proved using L’Hospital rule) shows that for the sequence xn ≈ ln2 n, the corresponding required subsequence is k(n) = n + [nc/(2ln n)] (c ≥ 0).
A particular case (with a = b =1) of sequences in my previous answer is the most famous sequence in Number Theory (and one of the most famous sequence in Mathematics) xn = π (n) ≈ n/ln n with the related required subsequence k(n) = n + [cln n] (n = 1,2,…; c ≥ 0). Namely, the Prime Number Theorem describes the asymptotic distribution of the prime numbers among the positive integers. More precisely, the Prime Number Theorem gives an asymptotic form for the prime counting function π (n), which counts the number of prime numbers less than some integer n, as follows:
I like your proposition, however with a feeling that it is just a conjecture. My doubts came from the reading of the Prime Numbers Theorem on their distribution which state only that the RATIO of pi (n)/ [ n / ln n ] possesses a positive limit.
Probably this is NOT sufficient for inferring your proposition from the true theorem about the sequence n /ln n.
You are right. It is well known in Number Theory that for the prime counting function π (n), π (n) ≈ n/ln n as n → ∞ (the asymptotic law of distribution of prime numbers), and π (n) - n/ln n > n/ln^2 n for all n ≥ 599. (see, e.g., https://en.wikipedia.org/wiki/Prime_number_theorem). Hence, π (n) - n/ln n → ∞ as n → ∞ (this is in fact, Sloane's OEIS sequence A057835, https://oeis.org/A057835).
However, motivated by the fact that for each fixed c ≥ 0, the subsequence xk(n) with k(n) = n + [cln n] of the sequence xn: = n/ln n (n = 1, 2, ...) satisfies the limit relation
xk(n) - xn → c as n → ∞,
it would be of interest to investigate the values of integer sequences
dn(c): = π(n + [cln n]) - π (n) (n = 1, 2, ...) for any fixed c ≥ 0.
Related investigations are well known as very hard problems in Analytic Number Theory concerning the asymptotic estimates of number of prime numbers in a short interval (especially, in our case in an interval I(n, c): = (n, n + [cln n]] which exactly contains π(n + [cln n]) - π (n) prime numbers).
My computations in Mathematica 9 show that for some fixed values c > 0, the values π(n + [cln n]) - π (n) are very close to c (and sometimes equal to c) for sufficiently large values n = n(c). This suggests the formulation of some “discrete version” of this problem proposed by George Stoica, concerning some classes of nondecreasing positive integer sequences that tend to infinity.
Fulfiling the estimate π (n) - n/ln n > n/ln^2 n for all n ≥ 599 jointly with the PNT also are not sufficient for deriving the following claim made by you: