Question

In determining the monotonicity of coefficients in a series expansion (which appeared in one of my study), I come across the following problem.

Let $p\ge 2$ be an integer, and $$6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i,~~i\ge0$$ with $d_0=d_1=d_2=1$. How to show $d_i>d_{i+1}$ for all $i\ge3$?

By easy calculation, $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}$. For a recursion of 2 or 3 terms, it is easy to proceed with induction, but what about 4 terms in this case?

Answer 1

I am a little bit upset that my previous answer to the question was downvoted by somebody. As far as I understand the idea of MO is not necessarily to produce final answers/solutions/responses but mostly to provide the ideas of approaching a given problem.

It's a question of time to solve this particular problem (I still wonder about where it comes from). Denote $$ F(x)=\sum_{i=0}^\infty d_ix^i=1+x+x^2+x^3+\dots $$ the generating function of the sequence. The given recursion can be then written as the differential equation of order 1, $$ \frac{F'}{F}=\frac{3p(2p^2+2p(p-1)x+(p-1)(2p-1)x^2)}{6p^3-6p^2x-3p(p-1)x^2-(p-1)(2p-1)x^3} $$ which can be explicitly solved: $$ F(x)=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p} $$ (one checks that the expansion indeed starts $1+x+x^2+x^3+\dots$).

The desired property $d_i>d_{i+1}$ for $i\ge3$ is equivalent to showing that the expansion $$ 1-(1-x)F(x)=c_0x^4+c_1x^5+c_2x^6+\dots $$ has all coefficients $c_0,c_1,c_2,\dots$ positive. In turn, the latter is equivalent to the positivity of all coefficients in the expansion $$ \frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)=4c_0x^3+5c_1x^4+6c_2x^5+\dots. $$ We have $$ \frac{d}{dx}\bigl(1-(1-x)F(x)\bigr) =\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p} $$ $$ -p(1-x)\biggl(\frac1p+\frac{(p-1)x}p+\frac{(p-1)(2p-1)x^2}{2p^3}\biggr) \cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1} $$ $$ =\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3 \cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1} $$ $$ =\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3 \cdot\Biggl(1+\sum_{n=1}^\infty\frac{(p+1)(p+2)\dots(p+n)}{n!} \biggl(\frac xp+\frac{(p-1)x^2}{2p^2}+\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^n\Biggr), $$ and the desired positivity follows.

Answer 2

Simple calculation shows $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}, 30p^3d_5=6p^2(4+p)d_4+(p-1)[3p(3+2p)+(2p-1)(2+3p)]d_3$, and so $d_3>d_4>d_5$.

Assume $d_i>d_{i+1}>d_{i+2}$ and $6p^3(i+2)d_{i+2}\ge 6p^2(i+1+p)d_{i+1}+(p-1)[3p(i+2p)+(2p-1)(i-1+3p)]d_i$ for all $i\ge3$ .

Then $6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+\frac{p-1}{p}[3p^2(i+1+2p)d_{i+1}+(2p-1)(i+3p)pd_i]$< $6p^2(i+2+p)d_{i+2}+\frac{p-1}{p} {6p^2(i+1+p)d_{i+1}+(p-1)[3p(i+2p)+(2p-1)(i-1+3p)]d_i }\le$ $6p^2(i+2+p)d_{i+2}+ \frac{p-1}{p} 6p^3(i+2)d_{i+2}=6p^3(i+3)d_{i+2}$, this shows $d_{i+3}< d_{i+2}$.

Moreover, $6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i>$ $6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_{i+1}=$ $6p^2(i+2+p)d_{i+2}+(p-1)[3p(i+1+2p)+(2p-1)(i+3p)]d_{i+1}$. This shows $6p^3(i+3)d_{i+3}\ge 6p^2(i+2+p)d_{i+2}+(p-1)[3p(i+1+2p)+(2p-1)(i+3p)]d_{i+1} $. Done.

Answer 3

For $p>1$, the characteristic polynomial of the recurrence, $f(x)=6p^3x^3-6p^2x^2-3p(p-1)x-(p-1)(2p-1)$, has a real zero $x_1$ between $0$ and $1$ (this is easy) and a pair of complex conjugate zeros $x_2$ and $x_3$ whose absolute value is strictly less than $|x_1|$ (this is less trivial). Then a generic solution $d_i$ of the recurrence satisfies $d_{i+1}/d_i\to x_1$ as $i\to\infty$, so it satisfies $d_i>d_{i+1}$ eventually. It is technically possible (by expecting the corresponding DE for the generating series, for example,--but this is in any case worth trying) to show that your solution is generic, that is, $d_i^{1/i}\to x_1$ as $i\to\infty$. Replacing "eventually" by an effective bound (an "effective" version of Poincare's theorem) is usually hard unless the asymptotics can be computed using a different representation of the sequence.