Some issues with proving that a sequence is convergent I recently tried (in the sense that I believe the thesis holds) to prove that, given $a\in\mathbb{R}^+$, there exists
$$\lim_{n\rightarrow+\infty}\sqrt[n]{\sum_{k=0}^{\lfloor n/5\rfloor}{n-4k\choose k}a^{-5k}}\in\mathbb{R}$$
For those who do not like the floor brackets, $\sum_{k=0}^{\lfloor n/5\rfloor}{n-4k\choose k}a^{-5k}=\sum_{k=0}^{+\infty}{n-4k\choose k}a^{-5k}$.
Or one can also observe that it is sufficient to prove it for the subsequence $(n_s:=5s)_{s\in\mathbb{N}}$, which amounts to prove the existence of
$$\lim_{s\rightarrow+\infty}\sqrt[s]{\sum_{k=0}^s{5s-4k\choose k}a^{-5k}}\in\mathbb{R}$$
While showing that
$$\sup_{n\in\mathbb{N}^+}\sqrt[n]{\sum_{k=0}^{\lfloor n/5\rfloor}{n-4k\choose k}a^{-5k}}\le 2\max(a,1)\sup_{n\in\mathbb{N}^+}\sqrt[n]{n}\le 4\max(a,1)$$
is pretty straight forward, I'm having several issues with showing that the series is convergent.
I tried computations with a couple of values of $a$ and not only there was convergence, but it seemed to be definitely monotone increasing.
My attempts to prove any of that were not particularly fruitful, though.
A useful thing to know is that it comes from a power series: indeed, $\sum_{k=0}^{+\infty}{n-4k\choose k}a^{-5k}$ is the $n$-th coefficient of the Taylor series at $0$ of $$\frac{1}{1-x-\left(\frac{x}{a}\right)^5}$$
So, if the thesis holds, the limit must be the inverse of the positive real root of $x^5+a^4x-a^5$.
If someone more skillful could solve it, I'd be overjoyed. Thank you in advance.
Added I also tried to prove that $\lim_n \left(\left(\frac{1}{c_n}\right)^5+\frac{a^4}{c_n}-a^5\right)=0$, but I could not go through. Maybe it's just me, though.
 A: 
A useful thing to know is that it comes from a power series: indeed, $\sum_{k=0}^{+\infty}{n-4k\choose k}a^{-5k}$ is the $n$-th coefficient of the Taylor series at $0$ of $$\frac{1}{1-x-\left(\frac{x}{a}\right)^5}$$

This is the key. Let $\zeta$ be the positive zero of $p(z) = 1 - z - \left(\frac{z}{a}\right)^5$. Then $\zeta$ is the zero of $p$ with the strictly smallest modulus, and it is a simple zero ($p'(t) \leqslant -1$ for $t\in [0,+\infty)$). Thus
$$g(z) = \frac{1}{1-z-\left(\frac{z}{a}\right)^5} - \frac{1}{p'(\zeta)(z-\zeta)}$$
is holomorphic in a disk with radius $r > \zeta$. Let the Taylor series of $g$ be
$$g(z) = \sum_{n=0}^\infty \gamma_n z^n.$$
Further we have
$$\frac{1}{z-\zeta} = -\sum_{n=0}^\infty \frac{z^n}{\zeta^{n+1}}.$$
Then we find
$$a_n := \sum_{k=0}^\infty \binom{n-4k}{k}a^{-5k} = \gamma_n - \frac{1}{p'(\zeta)\zeta^{n+1}}.$$
Taking $n$-th roots, we obtain
$$\sqrt[n]{a_n} = \frac{1}{\zeta}\sqrt[n]{\gamma_n\zeta^n - \frac{1}{p'(\zeta)\zeta}} = \zeta^{-1}\sqrt[n]{\frac{1}{\zeta\lvert p'(\zeta)\rvert}}\sqrt[n]{1- p'(\zeta)\gamma_n\zeta^{n+1}}.$$
But
$$\sqrt[n]{\frac{1}{\zeta\lvert p'(\zeta)\rvert}} \to 1,$$
and since the radius of convergence of $g$ is greater than $\zeta$, we have
$$\gamma_n \zeta^n \to 0,$$
whence
$$1- p'(\zeta)\gamma_n\zeta^{n+1} \to 1$$
and also
$$\sqrt[n]{1- p'(\zeta)\gamma_n\zeta^{n+1}} \to 1.$$
Thus $\sqrt[n]{a_n} \to \zeta^{-1}$.
