How to solve this limit problem? $\lim_{n \to \infty} \frac{n^k}{(1+b)^n} = 0$ For $k \in \mathbb{N}$ and $b > 0$ show that 
$\lim_{n \to \infty} \frac{n^k}{(1+b)^n} = 0$
 A: $$\lim_{n\to\infty}\frac{\frac{(n+1)^k}{(1+b)^{n+1}}}{\frac{n^k}{(1+b)^n}}=\frac{1}{1+b}<1$$
hence $\sum_{n} \frac{n^k}{(1+b)^n}<\infty$ converges, so $\frac{n^k}{(1+b)^n}\to 0$
A: Consider:
$$\sum_{n \geq 1} \frac{n^k}{(1+b)^n}$$
Show it converges for $b>0$ by the ratio test, hence we must have:
$$\lim_{n \to \infty} \frac{n^k}{(1+b)^n}=0$$
For $b>0$, because if we didn't than by the divergence test the series would diverge for $b>0$ but it doesn't.
A: For $n\geqslant2k$ we have 
$$
\begin{aligned}
(1+b)^n&>\binom{n}{k+1}\\\\&=\dfrac{n(n-1)\cdots(n-k)}{(k+1)!}\\\\&>\dfrac{n^{k+1}}{2^{k+1}(k+1)!}
\end{aligned}
$$
and hence $n^k/(1+b)^n<2^{k+1}(k+1)!/n$ and since the RHS of the inequality tends to zero as $n\to\infty,$ we are done.
A: Using $a^{n} = e^{n \, \ln(a)}$ then
\begin{align}
\lim_{n \to \infty} \frac{n^k}{(1+b)^n} &= \lim_{n \to \infty} n^k \, e^{- n \, \ln(1 + b)} \to 0
\end{align}
since $\lim_{n \to \infty} e^{-a \,n} \to 0$ faster than $\lim_{n \to \infty} n^{k} \to \infty$.
A: Hint: For $n\ge\frac1{\left(1+b\right)^{\frac1{2k}}-1}$
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=\frac{(n+1)^k(1+b)^n}{n^k(1+b)^{n+1}}\\
&=\frac{\left(1+\frac1n\right)^k}{1+b}\\[6pt]
&\le\frac1{\sqrt{1+b}}
\end{align}
$$
