Sequences & Series – Problem 2: Prove that is a square-recursive sequence, and that is a geometric sequence

If a sequence $\{a_n\}$ satisfies $a_{n+1}=a_n^2$, then $\{a_n\}$ is called a "square-recursive sequence". For the given sequence $\{a_n\}$, $a_1=8$, and the point $(a_n,a_{n+1})$ lies on the graph of $$f(x)=x^2+4x+2$$ where $n$ is a positive integer.

(1) Prove that $\{a_n+2\}$ is a square-recursive sequence, and that $\{\lg(a_n+2)\}$ is a geometric sequence.

(2) Let $b_n=\lg(a_n+2)$, $c_n=2n-1$, and $$d_n=\begin{cases} c_n,& n\text{ is odd},\\ b_n,& n\text{ is even}. \end{cases}$$ Let $S_n$ be the sum of the first $n$ terms of $\{d_n\}$. If $$S_{2n}-2n^2+n+10\ge\lambda b_n$$ holds for all $n$, find the maximum value of $\lambda$.

(1) Since the point $(a_n,a_{n+1})$ lies on the graph of $f(x)=x^2+4x+2$, $$a_{n+1}=a_n^2+4a_n+2,$$ so $$a_{n+1}+2=(a_n+2)^2,$$ therefore $\{a_n+2\}$ is a square-recursive sequence.

Also, $\lg(a_1+2)=\lg10=1$, and $$\lg(a_{n+1}+2)=2\lg(a_n+2),$$ so $\{\lg(a_n+2)\}$ is a geometric sequence with first term $1$ and common ratio $2$.

(2) From part (1), $b_n=\lg(a_n+2)=2^{n-1}$, $$d_n=\begin{cases} 2n-1,& n\text{ is odd},\\ 2^{n-1},& n\text{ is even}. \end{cases}$$ Hence $$S_{2n}=\sum_{k=1}^{n}(4k-3)+\sum_{k=1}^{n}2^{2k-1}=2n^2-n+\frac{2(4^n-1)}{3}.$$ From $S_{2n}-2n^2+n+10\ge\lambda b_n$, we get $$\lambda\le\frac{\frac{2(4^n-1)}{3}+10}{2^{n-1}}=\frac{2^{2n+1}+28}{3\cdot2^{n-1}}.$$ Let $t=2^{n-1}\ge1$. Then the right-hand side becomes $\dfrac{8t^2+28}{3t}$. Over the discrete values $t=1,2,4,\ldots$ (since $t=2^{n-1}$), evaluate: $$\frac{8(1)^2+28}{3(1)}=\frac{36}{3}=12,\qquad \frac{8(2)^2+28}{3(2)}=\frac{60}{6}=10,\qquad \frac{8(4)^2+28}{3(4)}=\frac{156}{12}=13.$$ The minimum is $10$, attained at $t=2$ (i.e. $n=2$).

Therefore $\lambda\le10$, and this value is attainable, so $\lambda_{\max}=10$.

(1) From $a_{n+1}+2=(a_n+2)^2$, the sequence $\{a_n+2\}$ is square-recursive. Taking logarithms gives $\lg(a_{n+1}+2)=2\lg(a_n+2)$, so $\{\lg(a_n+2)\}$ is geometric with first term $1$ and ratio $2$.

(2) Using $b_n=2^{n-1}$ and the odd-even definition of $d_n$, we obtain $$S_{2n}=2n^2-n+\frac{2(4^n-1)}{3}.$$ After rewriting the inequality, the largest constant that works for all $n$ is $\lambda_{\max}=10$.