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I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

later From a sequence $\mathbf{f}=(f_1,\cdots)$ (with an intial term $f_0$) one obtains a sequence $\mathbf{d}=(d_1,\cdots)$ of harmonic differences (the difference $d_0$ is defined but will always be 0). This is a linear transformation whose kernel is the constant sequences. We will thus assume that all terms have been shifted so that $f_0=0$. This, of course, does not change the growth rate . Then the transformation is invertible. View the sequences as column vectors and let $M$ be the infinite lower-triangular non-negative matrix with $M_{nk}=2^{-n}\binom{n}{k}$ for $n,k \ge 1$. Then $(I-M)\mathbf{f}=\mathbf{d}$. The question is:

$\bullet$ What can be infered about $\mathbf{f}$ given that $\lim_{n \to \infty} d_n=\ell$ exists and is finite?

Independent of the limit existing or not, $\mathbf{f}=(I-M)^{-1}\mathbf{d}$. Furthermore, $$(I-M)^{-1}=I+M+M^2+M^3+\cdots$$ In this sum the terms are all non-negative matrices. At this stage I run out of easy theoretical observations. Numerically many things seem apparent, some of which I can easily prove and others which I can't but others surely can (if they are true).Let $N=(I-M)^{-1}$. The diagonal entries are $N_{kk}=\frac{2^k}{2^k-1}$ while $N_{k,k-1}=\frac{k2^k}{(2^k-1)(2^k-2)}$. The typical row (not too close to the top) seems to start approximately $[1.44,0.721,0.480,0.361..$ so very nearly $q[1,1/2,1/3,1/4...$ for some constant $q$ I don't recognize. It would appear that the entries$N_{nk}$ for $k$ about $\sqrt{n}$ on either side of $n/2$ sum to about 1. There are other rather smaller bulges maybe centered on $n/4$, $n/8$ etc.

So (roughly speaking) given a row not too near the top, the entries start with 1.44 and drop off quickly, the final nonzero entries are very small until a 1 on the diagonal. The entries in between are relatively small and the most signifigant ones are in the middle and have a constant sum. Succesive rows are very nearly equal in all entries except the diagonal. More precise quantitative versions of this might suffice to show that if $\lim_{n \to \infty} d_n=0$ then (and only then?) $\mathbf{f}=N\mathbf{d}$ has a finite limit $\lim_{n \to \infty} f_n=L$. This would seem enough because then if $\lim_{n \to \infty} d_n=\ell$ the $\mathbf{f}$ corresponding to it has $f_n-\frac{\ell}{\log 2}H_n$ going to some finite limit depending on the initial terms of $\mathbf{d}$.

I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

later From a sequence $\mathbf{f}=(f_1,\cdots)$ (with an intial term $f_0$) one obtains a sequence $\mathbf{d}=(d_1,\cdots)$ of harmonic differences (the difference $d_0$ is defined but will always be 0). This is a linear transformation whose kernel is the constant sequences. We will thus assume that all terms have been shifted so that $f_0=0$. This, of course, does not change the growth rate . Then the transformation is invertible. View the sequences as column vectors and let $M$ be the infinite lower-triangular non-negative matrix with $M_{nk}=2^{-n}\binom{n}{k}$ for $n,k \ge 1$. Then $(I-M)\mathbf{f}=\mathbf{d}$. The question is:

$\bullet$ What can be infered about $\mathbf{f}$ given that $\lim_{n \to \infty} d_n=\ell$ exists and is finite?

Independent of the limit existing or not, $\mathbf{f}=(I-M)^{-1}\mathbf{d}$. Furthermore, $$(I-M)^{-1}=I+M+M^2+M^3+\cdots$$ In this sum the terms are all non-negative matrices. At this stage I run out of easy theoretical observations. Numerically many things seem apparent, some of which I can easily prove and others which I can't but others surely can (if they are true).Let $N=(I-M)^{-1}$. The diagonal entries are $N_{kk}=\frac{2^k}{2^k-1}$ while $N_{k,k-1}=k\frac{2^{k-1}}{(2^k-1)(2^{k-1}-1)}$ and $N_{k,k-2}=\binom{k}{2}\frac{(2^{k-1}+1)2^{k-2}}{(2^k-1)(2^{k-1}-1)(2^{k-2}-1)}$. The typical row (not too close to the top) seems to start approximately $[1.44,0.721,0.480,0.361..$ so very nearly $q[1,1/2,1/3,1/4...$ for some constant $q$ I don't recognize. It would appear that the entries$N_{nk}$ for $k$ about $\sqrt{n}$ on either side of $n/2$ sum to about 1. There are other rather smaller bulges maybe centered on $n/4$, $n/8$ etc.

So (roughly speaking) given a row not too near the top, the entries start with 1.44 and drop off quickly, the final nonzero entries are very small until a 1 on the diagonal. The entries in between are relatively small and the most signifigant ones are in the middle and have a constant sum. Succesive rows are very nearly equal in all entries except the diagonal. More precise quantitative versions of this might suffice to show that if $\lim_{n \to \infty} d_n=0$ then (and only then?) $\mathbf{f}=N\mathbf{d}$ has a finite limit $\lim_{n \to \infty} f_n=L$. This would seem enough because then if $\lim_{n \to \infty} d_n=\ell$ the $\mathbf{f}$ corresponding to it has $f_n-\frac{\ell}{\log 2}H_n$ going to some finite limit depending on the initial terms of $\mathbf{d}$.

I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

later From a sequence $\mathbf{f}=(f_1,\cdots)$ (with an intial term $f_0$) one obtains a sequence $\mathbf{d}=(d_1,\cdots)$ of harmonic differences (the difference $d_0$ is defined but will always be 0). This is a linear transformation whose kernel is the constant sequences. We will thus assume that all terms have been shifted so that $f_0=0$. This, of course, does not change the growth rate . Then the transformation is invertible. View the sequences as column vectors and let $M$ be the infinite lower-triangular non-negative matrix with $M_{nk}=2^{-n}\binom{n}{k}$ for $n,k \ge 1$. Then $(I-M)\mathbf{f}=\mathbf{d}$. The question is:

$\bullet$ What can be infered about $\mathbf{f}$ given that $\lim_{n \to \infty} d_n=\ell$ exists and is finite?

Independent of the limit existing or not, $\mathbf{f}=(I-M)^{-1}\mathbf{d}$. Furthermore, $$(I-M)^{-1}=I+M+M^2+M^3+\cdots$$ In this sum the terms are all non-negative matrices. At this stage I run out of easy theoretical observations. Numerically mainy things seem apparent, some of which I can easily prove and others which I can't but others surely can (if they are true).Let $N=(I-M)^{-1}$. The diagonal entries are $N_{kk}=\frac{2^k}{2^k-1}$ while $N_{k,k-1}=\frac{k2^k}{(2^k-1)(2^k-2)}$. The typical row (not too close to the top) seems to start approximately $[1.44,0.721,0.480,0.361..$ so very nearly $q[1,1/2,1/3,1/4...$ for some constant $q$ I don't recognize. It would appear that the entries$N_{nk}$ for $k$ about $\sqrt{n}$ on either side of $n/2$ sum to about 1. There are other rather smaller bulges maybe centered on $n/4$, $n/8$ etc.

So (roughly speaking) given a row not too near the top, the entries start with 1.44 and drop off quickly, the final nonzero entries are very small until a 1 on the diagonal. The entries in between are relatively small and the most signifigant ones are in the middle and have a constant sum. Succesive rows are very nearly equal in all entries except the diagonal. More precise quantitative versions of this might suffice to show that if $\lim_{n \to \infty} d_n=0$ then (and only then?) $\mathbf{f}=N\mathbf{d}$ has a finite limit $\lim_{n \to \infty} f_n=L$. This would seem enough because then if $\lim_{n \to \infty} d_n=\ell$ the $\mathbf{f}$ corresponding to it has $f_n-\frac{\ell}{\log 2}H_n$ going to some finite limit depending on the initial terms of $\mathbf{d}$.

I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

later From a sequence $\mathbf{f}=(f_1,\cdots)$ (with an intial term $f_0$) one obtains a sequence $\mathbf{d}=(d_1,\cdots)$ of harmonic differences (the difference $d_0$ is defined but will always be 0). This is a linear transformation whose kernel is the constant sequences. We will thus assume that all terms have been shifted so that $f_0=0$. This, of course, does not change the growth rate . Then the transformation is invertible. View the sequences as column vectors and let $M$ be the infinite lower-triangular non-negative matrix with $M_{nk}=2^{-n}\binom{n}{k}$ for $n,k \ge 1$. Then $(I-M)\mathbf{f}=\mathbf{d}$. The question is:

$\bullet$ What can be infered about $\mathbf{f}$ given that $\lim_{n \to \infty} d_n=\ell$ exists and is finite?

Independent of the limit existing or not, $\mathbf{f}=(I-M)^{-1}\mathbf{d}$. Furthermore, $$(I-M)^{-1}=I+M+M^2+M^3+\cdots$$ In this sum the terms are all non-negative matrices. At this stage I run out of easy theoretical observations. Numerically many things seem apparent, some of which I can easily prove and others which I can't but others surely can (if they are true).Let $N=(I-M)^{-1}$. The diagonal entries are $N_{kk}=\frac{2^k}{2^k-1}$ while $N_{k,k-1}=\frac{k2^k}{(2^k-1)(2^k-2)}$. The typical row (not too close to the top) seems to start approximately $[1.44,0.721,0.480,0.361..$ so very nearly $q[1,1/2,1/3,1/4...$ for some constant $q$ I don't recognize. It would appear that the entries$N_{nk}$ for $k$ about $\sqrt{n}$ on either side of $n/2$ sum to about 1. There are other rather smaller bulges maybe centered on $n/4$, $n/8$ etc.

So (roughly speaking) given a row not too near the top, the entries start with 1.44 and drop off quickly, the final nonzero entries are very small until a 1 on the diagonal. The entries in between are relatively small and the most signifigant ones are in the middle and have a constant sum. Succesive rows are very nearly equal in all entries except the diagonal. More precise quantitative versions of this might suffice to show that if $\lim_{n \to \infty} d_n=0$ then (and only then?) $\mathbf{f}=N\mathbf{d}$ has a finite limit $\lim_{n \to \infty} f_n=L$. This would seem enough because then if $\lim_{n \to \infty} d_n=\ell$ the $\mathbf{f}$ corresponding to it has $f_n-\frac{\ell}{\log 2}H_n$ going to some finite limit depending on the initial terms of $\mathbf{d}$.

I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

Another rather vague thought: If we have an $f()$ which is increasing, not very quickly, then almost all the weight in the difference involving $f(n)$ is concentrated around the middle $f(n/2)$ so one has something close to the sequences with $\lim g(n)-g(n/2)=L $. Those in turn might be expected to be like $g(n)=L \log_2(n)=log_b(n)$ for $b=\sqrt[L]2$.

I also agree that logarithmic growth looks like the right thing. There is lots of freedom, one can make the sequence of differences whatever one wishes. In case we want the sequence to start with a and from then on to have that difference always be exactly L one gets a sequence starting a, 2L+a, (8/3)L+a, (22/7)L+a, (368/105)L+a, (2470/651)L+a, (7880/1953)L+a, (150266/35433)L+a which appears to involve A158466 in the OEIS, an interesting sequence.

later From a sequence $\mathbf{f}=(f_1,\cdots)$ (with an intial term $f_0$) one obtains a sequence $\mathbf{d}=(d_1,\cdots)$ of harmonic differences (the difference $d_0$ is defined but will always be 0). This is a linear transformation whose kernel is the constant sequences. We will thus assume that all terms have been shifted so that $f_0=0$. This, of course, does not change the growth rate . Then the transformation is invertible. View the sequences as column vectors and let $M$ be the infinite lower-triangular non-negative matrix with $M_{nk}=2^{-n}\binom{n}{k}$ for $n,k \ge 1$. Then $(I-M)\mathbf{f}=\mathbf{d}$. The question is:

$\bullet$ What can be infered about $\mathbf{f}$ given that $\lim_{n \to \infty} d_n=\ell$ exists and is finite?

Independent of the limit existing or not, $\mathbf{f}=(I-M)^{-1}\mathbf{d}$. Furthermore, $$(I-M)^{-1}=I+M+M^2+M^3+\cdots$$ In this sum the terms are all non-negative matrices. At this stage I run out of easy theoretical observations. Numerically mainy things seem apparent, some of which I can easily prove and others which I can't but others surely can (if they are true).Let $N=(I-M)^{-1}$. The diagonal entries are $N_{kk}=\frac{2^k}{2^k-1}$ while $N_{k,k-1}=\frac{k2^k}{(2^k-1)(2^k-2)}$. The typical row (not too close to the top) seems to start approximately $[1.44,0.721,0.480,0.361..$ so very nearly $q[1,1/2,1/3,1/4...$ for some constant $q$ I don't recognize. It would appear that the entries$N_{nk}$ for $k$ about $\sqrt{n}$ on either side of $n/2$ sum to about 1. There are other rather smaller bulges maybe centered on $n/4$, $n/8$ etc.

So (roughly speaking) given a row not too near the top, the entries start with 1.44 and drop off quickly, the final nonzero entries are very small until a 1 on the diagonal. The entries in between are relatively small and the most signifigant ones are in the middle and have a constant sum. Succesive rows are very nearly equal in all entries except the diagonal. More precise quantitative versions of this might suffice to show that if $\lim_{n \to \infty} d_n=0$ then (and only then?) $\mathbf{f}=N\mathbf{d}$ has a finite limit $\lim_{n \to \infty} f_n=L$. This would seem enough because then if $\lim_{n \to \infty} d_n=\ell$ the $\mathbf{f}$ corresponding to it has $f_n-\frac{\ell}{\log 2}H_n$ going to some finite limit depending on the initial terms of $\mathbf{d}$.