Is there a sequence $ c_n $ such that $$ \lim\sup_{n\to\infty}(\frac{1+c_{n+1}}{c_n})^n \lt e $$?
I tried to get the above sequence into a form that would reveal more about the potential upper limit, but I couldn't really figure out anything useful. How do I tackle this sort of problem?
$c_n=-1$ for $n$ even and $c_n=-2$ for $n$ odd. Then $sup(\frac{1+c_{n+1}}{c_n})=1$ for $c_{n+1}=-2$ and $1^n\to 1<e$
Let $0 < a< 1$. Solving
$$\frac{1+c_{n+1}}{c_n}=a$$ yields $$c_{n+1}=ac_n-1$$
Define $c_1= \alpha$ and $c_{n+1}=ac_n-1$ , where $\alpha$ is well chosen (*).
Then by construction
$$\left( \frac{1+c_{n+1}}{c_n} \right)^n=a^n \,.$$ So your sup limit is actually a limit, and is $0$.
Now, there is something subtle which we have to be careful with, and here is where $(*)$ comes in play:
We have to make sure that $c_n \neq 0$ for all $n$, otherwise we don't have $\frac{1+a_{n+1}}{a_n}=a$. Using the recurrence, if $c_n=0$ you can calculate backwards $c_{n-1},..,c_1$ (or show that there is no value of $c_i$ which leads to $c_n=0$) . Thus, you can prove that for each $n$ there is at most a value of $\alpha$ such that $c_n=0$.
This shows that there are at most countably many values of $\alpha$ which make a $c_n=0$. Picking any of the remaining uncountably many reals will do the trick.
