One of my friend asked me following question.
Q: What is the maximum value of $\sum_{k=0}^\infty 2^{-k}\cdot \ln{C_k}$ ?
Here, the sequence of positive numbers $\{C_k\}$ satisfies following conditions.
(1) $x_{k+1}=\frac{1}{2}x_k-\sqrt{C_k}$
(2) $0 < C_k < \frac{x_k^2}{4}$
(3) $x_k>0$
In other words, $C_k$ is somehow measuring the difference of $x_{k+1}$ and $x_k$ which is another positive sequence. To maximize above infinite sum, which $C_k$ is optimal choice?
I tried using $e^{\sum_{k=0}^\infty 2^{-k}\cdot \ln{C_k}}=\Pi_{k=0}^\infty C_k\cdot e^{2^{-k}}$ and maximize this infinite multiplication, but it didnt work (at least for me)
So, I'm hoping someone can give me solution or hint. please help!
There is no maximum. To see this, consider any positive decreasing sequence $\{ x_k \}_{k \ge 0}$. Then $C_k$ satisfies your conditions (1),(2),(3) if and only if $$\begin{cases} x_{k+1}< \frac{1}{2}x_k \\ C_k = \left( \frac{1}{2}x_k - x_{k+1} \right) ^2\end{cases}$$
Now, suppose you have a maximum value $$MAX = \sum_k 2^{-k} \ln C_k$$ for a specific maximizing sequence $\{ x_k \}_{k \ge 0}$. You can easily see that picking a new sequence $$x'_k = 2 x_k$$ you have $$C'_k = \left( \frac{1}{2}x'_k - x'_{k+1} \right) ^2 = 4 \left( \frac{1}{2}x_k - x_{k+1} \right) ^2 = 4C_k$$ so that $$\sum_k 2^{-k} \ln C'_k = \sum_k 2^{-k} \ln 4 C_k = \sum_k 2^{-k} \ln C_k + \ln 4 \sum_k 2^{-k} > MAX + \ln 4$$ and you have a contradiction.
In other words, it is enough to consider a doubled sequence $\{ 2x_k \}_k$ to get a larger value of $\sum_k 2^{-k} \ln C_k$: this shows that there is no maximum.