Convergence and Limit of the infinite series

Question

How can I find the limit of the series

$$ \sum_{t=1}^{\infty} \frac{\ln(c+2^{t-1})}{2^t} \;\;\text{where}\; c > 0 $$

Since the terms are positive, and I have an intuition that $a_{n-1} \geq a_n$, the series converge to a limit point.

I tried to analytically show that the terms are decreasing, as below

$$ \frac{\ln(c+2^{n-1})}{2^n} \leq \frac{\ln(2^{n-1}+2^{n-1})}{2^n} \leq \frac{\ln 2}{2^{n-1}} $$

But I am not sure about the selection of the upper bound for the constant $c$. How can I show that the series converge and find the limit of it ?

Answer 1

Hint. As regards convergence note that $$\frac{a_{n+1}}{a_n}=\frac{\ln(c+2^{n})}{2^{n+1}}\cdot \frac{2^n}{\ln(c+2^{n-1})}= \frac{1}{2}\cdot \frac{\ln(c+2^{n})}{\ln(c+2^{n-1})}\sim \frac{1}{2}\cdot \frac{n\ln(2)}{(n-1)\ln(2)}\to \frac{1}{2}$$ where $a_n=\dfrac{\ln(c+2^{n-1})}{2^n}>0$. What may we conclude?

Answer 2

There is $N \in \mathbb N$ such that $a \le 2^n$ for all $n>N$. Thus

$0 \le a_n:=\frac{\ln(a+2^{n-1})}{2^n} \leq \frac{\ln(2^{n-1}+2^{n-1})}{2^n} =n\frac{\ln 2}{2^{n}}=:b_n$ for all $n>N$.

The ratio test shows that $ \sum b_n$ is convergent, hence $\sum a_n$ is convergent.