# What is maximum of $\sum_{k=0}^\infty 2^{-k}\cdot \ln{C_k}$

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)

• $C_k$ is defined in terms of $x_k$. What are $x_k$? – uniquesolution Jun 21 '17 at 14:44
• @ uniquesolution we don't know that. sequence $x_k$ is just positive real number sequence satisfying above three conditions. So, here is I'm wondering. The maximum value of above summation will depend on the initial value $x_0$? maybe not. because $C_k$ is depending on the difference of $x_{k+1}$ and $x_k$. So, maybe the maximum value is independent of $x_0$. But. that's just my guess... – KGEO Jun 21 '17 at 14:48
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.