<s>[INST] monotonicity from 4 term-recursion. 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?
 [/INST] 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.
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