Show that this sequence converges to $0$ The Question
For any fixed $k'\in\mathbb{N}$, for any $a\in\mathbb{R}^+$ and for any $n\in\mathbb{N}$, define the function $f:\mathbb{N}\to\mathbb{R}$ given by
\begin{equation*}
 x_n=f(n)=\frac{n^{k'}}{(1+a)^n}. 
 \end{equation*}
I want to show that $(x_n)$ converges to $0$.
For clarification, I don't include $0$ in $\mathbb{N}$.
Attempts at Solution
For any $\varepsilon>0$, I need to find a $N\in\mathbb{N}$ such that for any integer $n
\geq \mathbb{N}$, the following holds:
\begin{equation*}
 \left | \,\frac{n^{k'}}{(1+a)^n}-0 \, \right | <\varepsilon. 
 \end{equation*}
Since $x_m>0$ for any $m\in\mathbb{N}$, I can drop the absolute value signs and we get
\begin{equation*}
 \frac{n^{k'}}{(1+a)^n}<\varepsilon. 
 \end{equation*}
So, I considered $x^{k'}=\varepsilon(1+a)^x$ for any $x\in\mathbb{R}^+$. That equation does not have a closed form solution I think, so I will denote the bigger root of that as $x^*$. Now $N=\lceil x^* \rceil$ should be a candidate for the convergence definition. This is where I get stuck: how do I put that $N$ back to the convergence definition and show that $N$ really is a good candidate?
I wonder if there are any elegant proofs for this; mine is too ugly.
Also, I have a simple inequality that should play a role in this but I do not see how it fits. (My "proof" did not use the inequality)
For any fixed $k'\in\mathbb{N}$ and for any $n\geq k'$, consider $(1+a)^n$.
\begin{align*}
 (1+a)^n&=\sum_{k=0}^{n}\binom{n}{k}a^k1^{n-k}\\
 &=\sum_{k=0}^{k'-1}\binom{n}{k}a^k+\binom{n}{k'}a^{k'}+\sum_{k=k'+1}^{n}\binom{n}{k}a^k\\
 &>\sum_{k=0}^{k'-1}\binom{n}{k}0^k +\binom{n}{k'}a^{k'}+ \sum_{k=k'+1}^{n}\binom{n}{k}0^k\\
 &=\binom{n}{k'}a^{k'}. 
 \end{align*}
Thanks in advance!!
 A: Let me construct an elementary proof.
Let $\sqrt[2k']{1+a}=1+b$. Then, $b>0$ since $a>0$,
and
$$
x_n=\frac{n^{k'}}{(1+a)^n}=\frac{n^{k'}}{(1+b)^{2k'n}}=\left(\frac{\sqrt{n}}{(1+b)^{n}}{}\right)^{2k'}
$$
But
$$
(1+b)^n\ge 1+bn>bn,
$$
and thus
$$
\frac{1}{(1+b)^n}<\frac{1}{bn},
$$
and finally
$$
x_n=\left(\frac{\sqrt{n}}{(1+b)^{n}}{}\right)^{2k'}<\left(\frac{\sqrt{n}}{bn}\right)^{2k'}=b^{-2k'}\cdot\frac{1}{n^{k'}}
$$
Now, it suffices to show that the right hand side tends to zero, as $n$ tends to infinity.
A: This is a variation on the theme
$$
\lim_{x\to\infty}\frac{x^k}{e^x}=0 \tag{*}
$$
If you can prove that
$$
\lim_{y\to\infty}\frac{y^k}{(1+a)^y}=0\tag{**}
$$
you're done also with your sequence. Limits of functions are more flexible than limits of sequences; in this case you can observe that $(1+a)^y=e^{y\log(1+a)}$ and so transform the limit (**) into
$$
\lim_{x\to\infty}\frac{1}{(\log(1+a))^k}\frac{x^k}{e^x}
$$
with the substitution $x=y\log(1+a)$. The constant is irrelevant and so we just need to prove (*). With a further substitution $x=kz$, it becomes
$$
\lim_{z\to\infty}k^k\frac{z^k}{e^{kz}}=\lim_{z\to\infty}k^k\Bigl(\frac{z}{e^z}\Bigr)^{\!k}
$$
and we just have to show that
$$
\lim_{z\to\infty}\frac{z}{e^z}=0
$$
If you don't want to use l'Hôpital, you can observe that, for $z>0$,
$$
e^z>1+z+\frac{z^2}{2} \tag{***}
$$
(which can be proved with the mean value theorem) and therefore
$$
\frac{e^z}{z}>\frac{1}{z}+1+\frac{z}{2}
$$
Since the right-hand side has obviously limit $\infty$, we're done.
Proof of (***). Consider $f(z)=e^z-1-z-z^2/2$. Then $f(0)=0$ and $f'(z)=e^z-1-z$.
Now $f'(z)=0$ and $f''(z)=e^z-1$ which is positive for $z>0$. Hence $f'(z)>0$ for $z>0$ and consequently $f(z)>0$ for $z>0$.
A: You have
$$x_n=\frac{n^{k'}}{(1+a)^n}=\exp \left( k' \ln(n)-n \ln(1+a)\right) = \exp \left[ n \left( k' \frac{\ln(n)}{n}-\ln(1+a)\right)\right]  $$
Now, it is well known that
$$\lim_{n \rightarrow +\infty} \frac{\ln(n)}{n} =0$$
so $$\lim_{n \rightarrow +\infty} \left( k' \frac{\ln(n)}{n}-\ln(1+a)\right) = - \ln(1+a) <0$$
so$$\lim_{n \rightarrow +\infty} n\left( k' \frac{\ln(n)}{n}-\ln(1+a)\right) = - \infty$$
so
$$\lim_{n \rightarrow +\infty} x_n = 0$$
