Proving convergence/divergence of this series? $$\sum_{k=0}^{\infty}\frac{(k!)}{(k+1)!+2^k}$$
Anyone know how to show that this series converges or diverges? I've tried multiple tests and nothing seems to be working. Any help would be amazing!
 A: Since $(k+1)!$ grows faster than $2^k$ there exist some $K_0$ so that the denominator is equivalent to $(k+1)!$ $\forall k\geq K_0$ then:$\sum_0^\infty\frac{k!}{k+1!+2^k} \sim\sum_{K_0}^\infty \frac{1}{k+1}\rightarrow\infty$
Edit: I'll clarify my steps as suggested.
$(1)$ $k!\ge 2^k$ for big enough k
Proof: 
Consider $f_k=\dfrac{k!}{2^k}$; the Limit $\displaystyle{\lim_{k\rightarrow\infty}\dfrac{f_{k+1}}{f_k}=\lim_{k\to\infty}\dfrac{k+1}{2}=\infty};$
By the ratio test $\displaystyle{\lim_{k\to\infty}f_k=\infty}$
So by the precise definition of a limit there exist some $K_0\in\mathbb{N}$ so that for all $k\ge K_0\Rightarrow k!\ge2^k$ 
 We also get that $\displaystyle{\lim_{k\to\infty}\frac{1}{f_k}=\lim_{k\to\infty}\frac{2^k}{k!}=0}$
(2) "Let $a_k=\dfrac{k!}{(k+1)!+2^k}$and $b_k=\dfrac{k!}{(k+1)!}$ then if $\displaystyle{\lim_{k\to\infty}\frac{a_k}{b_k}=L, L\in(0,\infty)}$ by the limit comparision test the series associated to $a_k,b_k$ either both diverge or both converge."
Proof that the limit exist:
 $\displaystyle{\lim_{k\to\infty}\frac{a_k}{b_k}=\lim_{k\to\infty}\frac{(k+1)!}{(k+1)!+2^k}=\lim_{k\to\infty}\cfrac{1}{1+\dfrac{2^k}{(k+1)!}}=\dfrac{1}{\displaystyle{\lim_{k\to\infty}1+\frac{2^k}{(k+1)!}}}}$.
$\displaystyle{\lim_{k\to\infty}\frac{2^k}{(k+1)!}}=\lim_{k\to\infty}\frac{1}{k+1}\frac{2^k}{k!}=0\cdot0=0$
We get that $\displaystyle{\lim_{k\to\infty}\frac{a_k}{b_k}=1}$ 
So by the limit comparision test: 
$$\sum_{K_0}^{\infty}b_k=\sum_{K_0}^{\infty}\dfrac{1}{k+1}=\infty\Rightarrow \sum_{K_0}^{\infty}a_k=\infty$$ 
A: I suggest to divide the numerator and the denominator of the fraction by $k!$:
$$
\frac{k!}{(k+1)! + 2^k} = \frac{1}{k+1 + \frac{2^k}{k!}} \geq \frac{1}{k+3} ,
$$
since $\frac{2^k}{k!} \leq 2$ for all positive integer $k$.
Thus
$$
\sum_{k=0}^{\infty} \frac{k!}{(k+1)! + 2^k} \geq \sum_{k=0}^{\infty} \frac{1}{k+3} .
$$
