Show that this sequence converges to $0$

Question

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!!

Answer 1

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.

Answer 2

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 $z>0$.

Answer 3

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 $$\lim_{n \rightarrow +\infty} x_n = 0$$