Finding the limits $\lim_{n \to \infty} a^n/n!$ and $\lim_{n \to \infty} n^a/(1 + b)^n$ $$\lim_{ n\to\infty}\frac {a^n}{n!}$$ where $a$ belongs to the set $\Bbb R$
$$\lim_{n\to\infty}\frac{n^a}{(1+b)^n}$$ where $a$ belongs to the set $\Bbb R$ and $b$ belongs to $\Bbb R^+$.
 A: If $v_n=\dfrac{n^a}{(1+b)^n}$ then
$\dfrac{v_{n+1}}{v_n}=\dfrac{(n+1)^a\times (1+b)^n}{(1+b)^{n+1}\times n^a}=\left(1+\dfrac{1}{n}\right)^a\times \dfrac{1}{1+b}\to\dfrac{1}{1+b}<1$
Now use Ratio test.
For the other sequence recall $$e^a=\sum_{n=0}^\infty\dfrac{a^n}{n!}.$$
A: Hint: Let's $\mbox{sgn}:\mathbb{R}\to \{-1,0,+1\}$ a fuction 


*

*$\mbox{sgn}(a)=+1$ if $a>0$,

*$\mbox{sgn}(a)=\;\,0\;\,$  if $a=0$,

*$\mbox{sgn}(a)=-1$ if $a<0$.
Note that
$$
\frac{a^n}{n!}
=
\mbox{sgn}(a)^n\frac{|a|^n}{n!}
=
\mbox{sgn}(a)^n\prod_{k=1}^n\frac{|a|}{ k}
=
\mbox{sgn}(a)^n\left(\prod_{0\leq k\leq 2a}\frac{|a|}{ k}\right)\cdot \prod_{2a< k\leq n}\frac{|a|}{ k}
$$
implies
$$
-\left(\prod_{0\leq k\leq 2a}\frac{|a|}{ k}\right)\cdot \left( \frac{1}{2}\right)^n
\leq 
\mbox{sgn}(a)^n\left(\prod_{0\leq k\leq 2a}\frac{|a|}{ k}\right)\cdot \prod_{2a< k\leq n}\frac{|a|}{ k}
\leq
+\left(\prod_{0\leq k\leq 2a}\frac{|a|}{ k}\right)\cdot \left( \frac{1}{2}\right)^n
$$
by cause $2a< k $ implies $\frac{|a|}{k}<\frac{1}{2}$ and $\left(\prod_{2a\leq k\leq n}\frac{|a|}{ k}\right)<\left(\frac{1}{2}\right)^n$.
