Collection of intervals covers $[0,1]$?

Question

For each $n=1,2,3,...$ and each $m=0,1,2,...,n-1$, let $$K^n_m = \left[ \frac{3m-n+1}{n} , \frac{3m-n+2}{n} \right] \subset (-1,2). $$ I am struggling these with two questions for quite some time:

(1) $ \ $ The collection $\{ K^n_m \} \setminus \{ K^1_0 \}$ covers the interval $[0,1]$?

In case the answer for (1) is yes, then

(2) $ \ $ For each $ \ x \in [0,1] \ $ is there an infinite number of $ \, K^n_m \, $ such that $ \ x \in K^n_m \ $?

Answer 1

Yes to both (1) and (2).

There has been an overwhelming lack of response (either positive or negative) for the three days after I posted my answer (no votes, no comments, no nothing).

So I decided to make and post a picture, which would hopefully explain my solution.

$K^2$, $K^3$, $K^4$, etc. ">

One starts (with some difficulty for me) seeing some patterns after writing down in detail the partitions for enough many $n$, for me for $n=2$ up to $n=10$, though at hindsight up to $n=8$ would have sufficed. (Added up to $n=12$ later.)

I just realized I have not defined how I use the word "partition" in my answer. (My use is a bit unconventional since a partition usually means a disjoint cover, but I will define it and stick with it). By a partition I mean the collection that we get for a fixed $n$ when we let $m=0,1,2,...,n-1$ vary. For example,
the partition $K^2=\{[\frac{-1}2,0],[1,\frac32]\}=\{K^2_m:m=0,1\}$.
The partition $K^3=\{[\frac{-2}3,\frac{-1}3],[\frac13,\frac23],[\frac43,\frac52]\} =\{K^3_m:m=0,1,2\}$, and
the partition $K^4=\{[\frac{-3}4,\frac{-2}4],[\frac04,\frac14],[\frac34,\frac44],[\frac64,\frac74]\}=\{K^4_m:m=0,1,2,3\}$, etc.

So, for $n=2$ the partition is $\{[\frac{-1}2,0],[1,\frac32]\}$ (and the question is if the interval $[0,1]$ gets eventually covered).

For $n=4$, one element of the partition is $[0,\frac14]$. (When I say "one element of the partition" I mean there is some $m$ that goes with that given $n$. For $n=4$, if we take $m=1$ we get the interval $K^4_1=[0,\frac14]$ which is one element of the partition $K^4$.)
For $n=8$, one element of the partition is $[\frac28,\frac38]=[\frac14,\frac38]$. Note how the interval $[0,\frac14]$ (coming from $n=4$) together with $[\frac14,\frac38]$ (coming from $n=8$) have an endpoint in common, and taken together they cover $[0,\frac38]$. I didn't list all the intervals for $n=16$ but $[\frac6{16},\frac7{16}]=[\frac38,\frac7{16}]$ ought to be in there, let me verify. Indeed, with $m=7$ we get $[\frac{3m-n+1}{n},\frac{3m-n+2}{n}]=[\frac{3(7)-16+1}{16},\frac{3(7)-16+2}{16}]=[\frac{21-16+1}{16},\frac{21-16+2}{16}]=[\frac6{16},\frac7{16}]$
So $[\frac6{16},\frac7{16}]=[\frac38,\frac7{16}]$, together with $[0,\frac14]$ and $[\frac14,\frac38]$ (already considered earlier) cover $[0,\frac7{16}]$.

Note how the intervals $[0,\frac14]$, $[\frac14,\frac38]$, $[\frac38,\frac7{16}]$ etc (here formally $n=2\cdot2^k$, $k=1,2,3,...,$ and, for a particular suitable value of $m$ depending on $n$, $K^n_m=[\frac{2^{k-1}-1}{2^k},\frac{2^k-1}n]=[\frac{2^k-2}n,\frac{2^k-1}n]$ ) do not overlap, except at endpoints. They have lengths $\frac14,\frac18,\frac1{16}$, etc, respectively, so taken together they cover an interval with a left endpoint $0$ and of length $\frac14+\frac18+\frac1{16}+\cdots=\frac14\sum_{n=0}^\infty(\frac12)^n=\frac14(\frac1{1-\frac12})=\frac12$, that is they together cover the interval $[0,\frac12)$.
Remark. If we start with $k=0$, i.e. take $n=2\cdot2^k$, $k=0,1,2,3,...$, then in this manner we cover the interval $[\frac{-1}2,\frac12)$ (see the picture).

In a similar way (for $n=4,8,16,...$, which could be written as $n=2\cdot2^k, k=1,2,3,...$) one covers the interval $(\frac12,1]$ so together with $[0,\frac12)$ we have covered $[0,1]$, except for $\frac12$, which I need to take care of later.
Remark. If we start with $k=0$, i.e. take $n=2\cdot2^k$, $k=0,1,2,3,...$, then in this manner we cover the interval $(\frac12,\frac32]$.

To answer (2), take $n=3\cdot2^k, k=0,1,2,3,...$, and more generally $n=p\cdot2^k, k=0,1,2,3,...$, and note that for each prime $p$ we get a collection covering $[0,1]$ (we could use some self-similarity, translation and contraction by suitable factors to argue that this is indeed the case), and then also use that if $p$ and $q$ are different primes then $p\cdot2^k\not=q\cdot2^j$ for any non-negative integers $k,j$.
I mean that the collection $\{\cup K^n: k=0,1,2,3,...,\ n=2\cdot2^k\}$ covers $[0,1]$ except for the "problematic point" $\frac12$. (Here $\cup K^n=K^n_0\cup K^n_1\cup\cdots\cup K^n_{n-1}$, and the above collection covers $(-1,\frac12)\cup(\frac12,2)$, so $\frac12$ is not yet covered, see the picture.) Similarly, if $p$ is an odd prime, then the collection $\{\cup K^n: k=0,1,2,3,...,\ n=p\cdot2^k\}$ covers $[0,1]$, and covers $(-1,2)$, except for finitely many "problematic points" which I discuss next. (It could be easily seen that for each $n$ the problematic points are all $\frac l n$ that are not an endpoint of $K^n_m$ for any $m$, and where $-n<l<2n$ with $l$ an integer.)

The problematic point $\frac12$ (for prime $p=2$) is a rational number, and so is every "problematic point" for other primes $p$. So one could supply a separate argument that every rational number in $(-1,2)$ is covered infinitely many times. Say $r=\frac a b$, and for convenience I would also assume $r\in(0,1)$ so $0<a<b$ for some integers $a,b$. Then take $n=b$. If we are lucky, then $\frac a b$ is an endpoint of an interval in the partition that goes with $n=b$. Otherwise look at $n=b+1$ and (the reader may verify) that the interval $[\frac a {b+1},\frac {a+1} {b+1}]$ is in that partition. Since $\frac a {b+1}< \frac a b < \frac {a+1} {b+1}$ we have $\frac a b$ covered when $n=b+1$. Since I didn't require that the fraction $\frac a b$ be in lowest terms, we also get that every rational number in $(0,1)$ is covered infinitely many times (and one could extend this to show every rational in $(-1,2)$ is covered infinitely many times too). Here are some extra details. The number $0$ is an endpoint of the interval $[\frac{-1}n,\frac0n]\in K^n$ and the number $1$ is an endpoint of the interval $[\frac n n,\frac{n+1}n]\in K^n$ whenever $n=3j+2$, $j\ge0$. Similarly, $0$ is an endpoint of $[\frac0n,\frac1n]\in K^n$ and $1$ is an endpoint of $[\frac{n-1}n,\frac n n]\in K^n$ whenever $n=3j+1$, $j\ge1$. Also, $\cup_{k\ge0}[\frac{n+2k}{n+k},\frac{n+2k+1}{n+k}]=(1,2)$ whenever $n=3j+2$, $j\ge0$ (where the intervals involved belong to $K^{n+k}$ and each occurs only once). For example, for $n=2$ we have $(1,2)=[\frac22,\frac32]\cup[\frac43,\frac53]\cup[\frac64,\frac74]\cup\cdots$, and for $n=5$ we have $(1,2)=[\frac55,\frac65]\cup[\frac76,\frac86]\cup[\frac97,\frac{10}7]\cup\cdots$, etc, see the picture.) The picture in the interval $(-1,0)$ is symmetric about $\frac12$ to the picture in the interval $(1,2)$.

Just to illustrate and clarify what is meant by "problematic points". As above, for any given prime $p$ take $n=p\cdot2^k, k=0,1,2,3,...$. For example when $p=5$ then when $k=0$ we get $n=5$ and $[\frac{-1}5,\frac05],[\frac25,\frac35],[\frac55,\frac65]$ are some elements of this partition. So we need to cover $[\frac05,\frac25]$ and $[\frac35,\frac55]$. Let us focus on $[\frac05,\frac25]$, and next look at $k=1$ so $n=10$. Note that $[\frac0{10},\frac1{10}]$ and $[\frac3{10},\frac4{10}]$ are in that partition, so will will eventually cover $[\frac05,\frac25]$ except for its midpoint, which is problematic, $\frac15$. Similarly we will cover the interval $[\frac35,\frac55]$ except for the problematic (mid)point $\frac45$. Altogether, for $p=5$ we have two problematic points $\frac15$ and $\frac45$, both rational, and both covered when $n=6$ by $[\frac16,\frac26]$ and $[\frac46,\frac56]$ respectively.

The comment about self-similarity used above may be made more precise by the following lemma.
Lemma. Suppose that $t,q$ are integers (with $q$ positive) such that the intervals $[\frac{t-2}q,\frac{t-1}q]$ and $[\frac{t+1}q,\frac{t+2}q]$ belong to the partition $K^q$. Then the partitions $K^n$ for $n=q\cdot2^k,k=0,1,2,3,...$ taken together cover the interval $[\frac{t-2}q,\frac{t+2}q]$ except for the point $\frac{t}q$. (Besides, of course, they may cover some other intervals too.) There is also an easy to state condition telling us if an interval $[\frac{h}q,\frac{h+1}q]$ belongs to $K^q$ or not (assuming $-q<h$ and $h+1<2q$). The interval $[\frac{h}q,\frac{h+1}q]$ belongs to $K^q$ if (and only if) $q+h$ gives remainder $1$ when divided by $3$. In the above $q$ need not be a prime, but if $p$ and $q$ are different primes then the family (or what I called a partition) $K^{(p\cdot2^k)}$ has no elements in common with the family $K^{(q\cdot2^j)}$, for any non-negative integers $k,j$.

Edit September 1, 2019, to address the first comment below, to my answer. I will use (and prove later further down) that $[\frac{h}q,\frac{h+1}q]$ belongs to $K^q$ if and only if $q+h$ gives remainder $1$ when divided by $3$ (assuming, as in the preceding paragraph, that $-q<h$ and $h+1<2q$).
This is actually three cases:
Case (a), $h$ and $q$ mod $3$ are $1$ and $0$, respectively, or
Case (b), $h$ and $q$ mod $3$ are $0$ and $1$, respectively, or
Case (c), $h$ and $q$ mod $3$ are $2$ and $2$ respectively.
Consider case (a) and consider the interval $[\frac{2(h+1)}{2q},\frac{2(h+1)+1}{2q}]$ (note its left endpoint is the same as the right endpoint of $[\frac{h}q,\frac{h+1}q]$, but it is twice as short). In case (a) we have that $h+1$ is $2$ mod $3$, hence $2(h+1)$ and $2q$ are $1$ and $0$, respectively, mod $3$. Se we remain in case (a), that is the interval $[\frac{2(h+1)}{2q},\frac{2(h+1)+1}{2q}]$ matches case (a), and we may continue by induction. We "keep going to the right" with the length of the interval becoming twice as small at each step, and with the right endpoint of $[\frac{h}q,\frac{h+1}q]$ being the same as the left endpoint of the "next" interval $[\frac{2(h+1)}{2q},\frac{2(h+1)+1}{2q}]$.

Something similar happens if we "go to the left", that is from the interval $[\frac{h}q,\frac{h+1}q]$ we go to the interval $[\frac{2h-1}{2q},\frac{2h}{2q}]$. In case (a), we will have $2h-1$ and $2q$ are $1$ and $0$, respectively, mod $3$, so we remain in case (a). The left endpoint $\frac{h}q$ of the bigger interval is the same as the right endpoint $\frac{2h}{2q}$ of the smaller interval.

Now consider case (b). Before the formal proof, what we will get is that cases (b) and (c) alternate. I will consider the case when we "go to the left" (instead of going to the right, which could be considered in a similar way). So, if the interval $[\frac{h}q,\frac{h+1}q]$ matches case (b), consider the interval $[\frac{2h-1}{2q},\frac{2h}{2q}]$. Its right endpoint is the same as the left endpoint of $[\frac{h}q,\frac{h+1}q]$, but $[\frac{2h-1}{2q},\frac{2h}{2q}]$ is twice as short. Since, in case (b), $h$ and $q$ are $0$ and $1$ mod $3$, we have that $2h-1$ and $2q$ are $2$ and $2$ mod $3$, respectively, so $[\frac{2h-1}{2q},\frac{2h}{2q}]$ matches case (c). From case (b) "going to the left" we end up in case (c).

Now assume that the interval $[\frac{h}q,\frac{h+1}q]$ matches case (c), and go "to the left" again, taking the interval $[\frac{2h-1}{2q},\frac{2h}{2q}]$ which is twice as short. Since $[\frac{h}q,\frac{h+1}q]$ matches case (c), $h$ and $q$ are $2$ and $2$ mod $3$, thus we have that $2h-1$ and $2q$ are $0$ and $1$ mod $3$, respectively, so $[\frac{2h-1}{2q},\frac{2h}{2q}]$ matches case (b). From case (c) "going to the left" we end up in case (b). Thus, cases (b) and (c) alternate when "we go to the left" and intervals get twice as short at each step, with overlapping endpoints.

What was covered above was case (a) "going to the right", also case (a) "going to the left", and finally cases (b) and (c) together "going to the left". The remaining case is cases (b) and (c) taken together (alternating) and "going to the right", and it could be discussed in a similar way as above.

I have avoided the description in terms $\frac{3m-n+1}{n}$. I have used an alternative description, you pick an interval (i.e. pick some endpoints), and just from its endpoints you could tell if it is some $K^n_m$, here it is. We have that $K^n_m=[\frac{h}q,\frac{h+1}q]$ for some positive $n,q$ some $m,h$ (all integers) with $0\le m\le n-1$ and $-q<h<h+1<2q$ if and only if $n=q$ (note that $K^n_m$ has length $\frac1n$ while $[\frac{h}q,\frac{h+1}q]$ has length $\frac1q$)
and one of these three cases holds:
(a) if $q$ is $0$ mod $3$ then $h$ is $1$ mod $3$, or
(b) if $q$ is $1$ mod $3$ then $h$ is $0$ mod $3$, or
(c) if $q$ is $2$ mod $3$ then $h$ is $2$ mod $3$.
For example, if we start with $n,m$ and if $n$ is $2$ mod $3$, then we are in case (c), indeed mod $3$ we have $h=3m-n+1=-n+1=-2+1=-1=2$ (and $q=n=2$ mod $3$). Similarly when $n$ is $0$ or $1$ mod $3$. If one has to go the opposite way, start with $[\frac{h}q,\frac{h+1}q]$ with $q>0$ and $-q<h<h+1<2q$ then let $n=q$, so $h=3m-n+1$. Solving for $m$ we have $m=\frac{h+n-1}3$. Note (a),(b),and (c) (taken together) could be expressed shortly by $h+q=h+n=1$ mod $3$. So $h+n-1$ is divisible by $3$ and the formula $m=\frac{h+n-1}3$ produces an integer $m$. Since $-n=-q<h$ we have $h+n-1>-n+n-1=-1$, so $h+n-1\ge0$ (since it is an integer $>-1$) and hence $m\ge0$. Since $h<2q-1=2n-1$ we have $h+n-1<2n-1+n-1=3n-2$, hence $h+n-1\le3n-3$ (since $h+n-1$ is an integer $<3n-2$), hence $m\le\frac{3n-3}3=n-1$.

Answer 2

Yes to both (1) and (2).

Lemma (double $n$). Start with $K^n_m=[\frac{3m-n+1}{n} , \frac{3m-n+2}{n}] \subset (-1,2)$, with $0\le m\le n-1$. Clearly this interval has length $\frac1n$. Let $n'=n''=2n$, let $m'=2m$, let $m''=2m+1$. Then the intervals $K^{n'}_{m'} = \left[ \frac{3m'-n'+1}{n'} , \frac{3m'-n'+2}{n'} \right]$ and $K^{n''}_{m''} = \left[ \frac{3m''-n''+1}{n''} , \frac{3m''-n''+2}{n'} \right]$ each has length $\frac1{2n}$, also $\frac{3m'-n'+2}{n'}=\frac{3m-n+1}{n}$ and $\frac{3m-n+2}{n}=\frac{3m''-n''+1}{n''}$. That is, the left endpoint of $K^n_m$ is the right endpoint of $K^{n'}_{m'}$, and the right endpoint of $K^n_m$ is the left endpoint of $K^{n''}_{m''}$. Also, $0\le m'<m''\le n''-1=n'-1$.
Proof. Easy verification, left to the reader.

Lemma (going to the left). Start again with $K^n_m=[\frac{3m-n+1}{n} , \frac{3m-n+2}{n}]$, with $0\le m\le n-1$. Let $n(k)=n\cdot2^k$ and $m'(k)=m\cdot2^k$, $k=0,1,2,3,...$. Let $L^n_m=\cup_{k\ge0}K^{n(k)}_{m'(k)}$. Then $L^n_m=(\frac{3m-n}{n} , \frac{3m-n+2}{n}]$, an interval that does not contain its left endpoint, but contains its right endpoint which is the same as the right endpoint of $K^n_m$, and the length of $L^n_m$ is $\frac2n$ which is twice the length of $K^n_m$.
Proof. Easy verification (use the preceding lemma and use sum of geometric series with ratio $\frac12$), left to the reader. (The main idea, we start with $K^n_m$ and put intervals next to it, going to the left, overlapping only at endpoints, and the next interval to the left is always twice as short as its neighbor to which it was attached.)

Lemma (going to the right). Start again with $K^n_m=[\frac{3m-n+1}{n} , \frac{3m-n+2}{n}]$, with $0\le m\le n-1$. Let $n(k)=n\cdot2^k$, $m''(0)=m$ and $m''(k+1)=2m''(k)+1$, $k=0,1,2,3,...$. Let $R^n_m=\cup_{k\ge0}K^{n(k)}_{m''(k)}$. Then $R^n_m=[\frac{3m-n+1}{n},\frac{3m-n+3}{n})$, an interval that does not contain its right endpoint, but contains its left endpoint which is the same as the left endpoint of $K^n_m$, and the length of $R^n_m$ is twice the length of $K^n_m$.
Proof. Easy verification (similar to the proof of the lemma "going to the left"), left right to the reader.

Lemma (double $n$ to $\infty$). Start with $K^n_m$, and define $n(k)$, $m'(k)$, $m''(k)$, for $k=0,1,2,3,...$, as above.
Then the intervals $K^{n(k)}_{m'(k)}$ and $K^{n(k)}_{m''(k)}$ cover the open interval $D^n_m=L^n_m\cup R^n_m=(\frac{3m-n}{n},\frac{3m-n+3}{n})$.
That is, $\cup_{k=0}^{\infty}\bigl(K^{n(k)}_{m'(k)}\cup K^{n(k)}_{m''(k)}\bigr)=(\frac{3m-n}{n},\frac{3m-n+3}{n})$.
Proof. Left to the reader (follows directly from the preceding two lemmas).

Lemma (double $n$ to $\infty$, vary $m$). Start with $n$, let $n(k)=n\cdot2^k$ for $k=0,1,2,3,...$, and for each $n(k)$ take $K^{n(k)}_m$ for all possible $m$ allowed (i.e. $0\le m\le 2n(k)-1$). Then the union over $k$ together with union over all possible $m$ for each $k$ of the intervals $K^{n(k)}_m$ covers the interval $(-1,2)$ except for points of the form $\frac{3m-n}{n}$, $1\le m\le n-1$. (Note that if $m=0$ or $m=n$ in $\frac{3m-n}{n}$ then we get the points $-1$ and $2$ which are not considered anyway.)
Proof. Left to the reader, use preceding lemmas. The union in question could be written as $\cup_{k=0}^{\infty}\cup_{m=0}^{2n(k)-1}K^{n(k)}_m$. It is also used, of course, that the right endpoint of the interval $(\frac{3m-n}{n},\frac{3m-n+3}{n})$ is the same as the left endpoint of the interval $(\frac{3(m+1)-n}{n},\frac{3(m+1)-n+3}{n})$ (and, if that wasn't enough, the left endpoint of the interval $(\frac{3m-n}{n},\frac{3m-n+3}{n})$ is the same as the right endpoint of the interval $(\frac{3(m-1)-n}{n},\frac{3(m-1)-n+3}{n})$, under obvious restrictions for $m$).

Lemma (arbitrarily large $n$). If $x\in(-1,2)$ and if $x$ is not of the form $\frac{3m-n}{n}$ for any $n$ and any $m$ with $1\le m\le n-1$ then $x$ is covered infinitely many times.
Proof. For any arbitrary large $n$ there is some $n(k)=n\cdot2^k\ge n$ (where $k\ge0$) and some $m$ with $0\le m\le n(k)-1$ such that $x\in K^{n(k)}_m=K^{n\cdot2^k}_m$. (Follows from the preceding lemma.)

Lemma (cover $\frac{3m-n}{n}$). For each $n$ and each $m$ with $1\le m\le n-1$ the point $\frac{3m-n}{n}$ is covered infinitely many times.
Proof. This lemma is true as stated, but for brevity it will only be proved under the additional assumption that $\frac{3m-n}{n}\in[0,1]$, that is $0\le3m-n\le n$, or equivalently $n\le3m\le2n$. (This is enough to answer the OP. The proof when $\frac{3m-n}{n}$ is in the interval $[1,2)$ is not difficult but goes in a different way (included at the end), and the proof for $(-1,0]$ is similar to the proof for $[1,2)$.)
For each $j\ge1$ (and assuming $0\le3m-n\le n$) we have $\frac{3m-n}{n}=\frac{3jm-jn}{jn}\in[\frac{3jm-(jn+1)+1}{jn+1},\frac{3jm-(jn+1)+2}{jn+1}]= K^{jn+1}_{jm}$. Indeed clearly $\frac{3jm-(jn+1)+1}{jn+1}=\frac{3jm-jn}{jn+1}\le\frac{3jm-jn}{jn}$ (using that these fractions are non-negative, and with equality only when the numerator is $0$). Also, the inequality $\frac{3m-n}{n}\le\frac{3jm-(jn+1)+2}{jn+1}$ is equivalent to $(3m-n)\cdot(jn+1)\le n\cdot(3jm-jn+1)$ which in turn is equivalent to $3m-n\le n$, which we assumed.

Lemma (bonus, alternative proof for $x\in[1,2)$ regardless whether it is of the form $\frac{3m-n}{n}$ or not).
Suppose $x\in[1,2)$ and $k\ge0$. Note that $\lim_{j\to\infty}\frac{3k+2+2j}{3k+2+j}=2$. There is a smallest $j\ge0$ such that $\frac{3k+2+2j+1}{3k+2+j}>x$. Then $x\in K^n_m$ where $n=3k+2+j$ and $m=2k+j+1$. (Note here we are free to start with larger and larger $k$, so $x$ is covered infinitely many times.)
Proof. We need to show that $x\ge\frac{3\cdot(2k+j+1)-(3k+2+j)+1}{3k+2+j}=\frac{3k+2+2j}{3k+2+j}$. Assume to the contrary that $x<\frac{3k+2+2j}{3k+2+j}$. Then $j\ge1$ for $j=0$ would result in the impossible $x<1$. Using the inequality $\frac a b<\frac{a-1}{b-1}$ which is easily verified assuming $a>b>1$, we get $x<\frac{3k+2+2j-1}{3k+2+j-1}=\frac{3k+2+2(j-1)+1}{3k+2+(j-1)}$ contradicting the minimal choice of $j$. Thus $x\ge\frac{3k+2+2j}{3k+2+j}$. We also need to show that $x\le\frac{3\cdot(2k+j+1)-(3k+2+j)+2}{3k+2+j}=\frac{3k+2+2j+1}{3k+2+j}$. But this is already satisfied by construction (the choice of $j$) we have $x<\frac{3k+2+2j+1}{3k+2+j}$. (Alternatively, note that when $j=0$ we start with the interval $[\frac{3k+2+2j}{3k+2+j},\frac{3k+2+2j+1}{3k+2+j}]=[1,1+\frac1{3k+2}]$, and when we go from $j$ to $j+1$ then the respective intervals overlap, eventually approaching $2$ from the left, as $j\to\infty$ (even if the length of these intervals goes to $0$). Indeed, $\frac{3k+2+2(j+1)}{3k+2+(j+1)}<\frac{3k+2+2(j+1)-1}{3k+2+(j+1)-1}=\frac{3k+2+2j+1}{3k+2+j}$, where we used again $\frac a b<\frac{a-1}{b-1}$ if $a>b>1$.)