Show that $\lim\limits_{t\to\infty}\frac{1-F(\eta t)}{1-F(t)}=0\implies E[X^m]<\infty$ 
Let $X\ge0$ be a random variable with distribution function $F(t)$ such that $F(t)<1$ for all $t\in\mathbb{R}$ and, for some $\eta\in (1,\infty)$,
\begin{align*}
\lim_{t\to\infty}\frac{1-F(\eta t)}{1-F(t)}=0.
\end{align*}
Show that $E[X^m]<\infty$ for any $m\in(0,\infty)$.

I have shown that $EX<\infty$, as done below:
Note that our assumption is that: $\lim\limits_{t\to\infty}\frac{P(X>\eta t)}{P(X>t)}\stackrel{(*)}{=}0$, we first show that $(*)\implies E[X]<\infty$. By $(*)$ $\exists$ an $s\in\mathbb{N}$ such that:
\begin{align*}
\frac{P(X>\eta t)}{P(X>t)}<\frac{1}{2\eta}\,\,\text{for all $t\ge s$}
\end{align*}
Now note that:
\begin{align*}
\int_{s\eta^n}^{s\eta^{n+1}}P(X>t)\,dt&\le P(X>s\eta^n)(s\eta^{n+1}-s\eta^{n})\\
&=s\eta^n(\eta-1)P(X>s\eta^n)\\
&=s\eta^n(\eta-1)\frac{P(X>s\eta^{n})}{P(X>s\eta^{n-1})}\frac{P(X>s\eta^{n-1})}{P(X>s\eta^{n-2})}...\frac{P(X>s\eta)}{P(X>s)}P(X>s)\\
&\le s\eta^n(\eta-1)\frac{1}{2\eta}\frac{1}{2\eta}...\frac{1}{2\eta}P(X>s)\,\,\text{since:}\\
&\text{$\bigg|\frac{s\eta^{n-k}}{s\eta^{n-k-1}}\bigg|=\eta$ and $s\eta^{n-k}=\eta(s\eta^{n-k-1})$ where $s\eta^{n-k-1}\ge s$ as $\eta>1$}\\
&\le s(\eta-1)\eta^n\frac{1}{(2\eta)^n}\quad\text{as $P(X>s)\le1$}\\
&=\frac{s(\eta-1)}{2^n}\\
\end{align*}
Thus,
\begin{align*}
&\int_{s\eta}^{\infty}P(X>t)\,dt=\sum_{n\ge1}\int_{s\eta^n}^{s\eta^{n+1}}P(X>t)\,dt\le s(\eta-1)\sum_{n\ge1}\frac{1}{2^n}=s(\eta-1)<\infty
\end{align*}
Hence,
\begin{align*}
EX=\int_{0}^{\infty}P(X>t)\,dt=\int_{0}^{s\eta}P(X>t)\,dt+\int_{s\eta}^{\infty}P(X>t)\,dt\le s\eta+s(\eta-1)<\infty\,\,\text{as we wished to show}.
\end{align*}
However, I cannot figure out how to extend this result to $EX^m<\infty$, any help here would be greatly appreciated. My thoughts on an extension are as follows, if we can show that
\begin{align*}
\lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}=0
\end{align*}
Then replacing $X$ with $X^m$ in our above argument finishes the proof, but I cannot show that this limit is zero. Here is what I have
\begin{align*}
\lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}&=\lim_{t\to\infty}\frac{P(X>(\eta t)^{1/m})}{P(X>t^{1/m})}\\
&=\lim_{t\to\infty}\frac{P(X>{\eta}^{1/m} t^{1/m})}{P(X>t^{1/m})}\\
&=\lim_{z\to\infty}\frac{P(X>{\eta}^{1/m}\cdot z)}{P(X>z)}\quad\text{since $z=t^{1/m}\to\infty$ as $t\to\infty$}
\end{align*}
But now $\eta>1$ implies that $\eta^{1/m}<\eta$ and so
\begin{align}
X>\eta z\implies X>\eta^{1/m}z
\end{align}
And so $P(X>\eta z)\le P(X>\eta^{1/m}z)$, hence
\begin{align*}
\lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}=\lim_{z\to\infty}\frac{P(X>{\eta}^{1/m}\cdot z)}{P(X>z)}\ge \lim_{z\to\infty}\frac{P(X>\eta z)}{P(X>z)}=0
\end{align*}
and so the inequality is going to the wrong way.
 A: Only slight modifications will be needed for the general case $m\in(0,\infty)$.
First, a change of variables $u\equiv t^{1/m}$ implies that
\begin{align*}
\mathbb E[X^m]=\int_0^{\infty}\mathbb P[X^m>t]\,\mathrm dt=\int_0^{\infty}\mathbb P[X>t^{1/m}]\,\mathrm dt=\int_0^{\infty}\mathbb P[X>u]m u^{m-1}\,\mathrm du.
\end{align*}
Second, take $s>0$ so large that
\begin{align*}
\frac{\mathbb P[X>\eta t]}{\mathbb P[X>t]}<\frac{1}{2\eta^m}\quad\text{for all $t\geq s$.}
\end{align*}
Third, for any $n\in\{0,1,2,\ldots\}$,
\begin{align*}
\int_{s\eta^n}^{s\eta^{n+1}}\mathbb P[X>u]m u^{m-1}\,\mathrm du&\leq\mathbb P[X>s\eta^n]\int_{s\eta^n}^{s\eta^{n+1}}m u^{m-1}\,\mathrm du\\
&=\mathbb P[X>s\eta^n]\left[(s\eta^{n+1})^m-(s\eta^{n})^m\right]\\
&=\mathbb P[X>s\eta^n]s^m\eta^{nm}(\eta^m-1)\\
&\leq\frac{\mathbb P[X>s]}{2^n\eta^{nm}}s^m\eta^{nm}(\eta^m-1)\\
&=\mathbb P[X>s]\frac{s^m(\eta^m-1)}{2^n}\\
&\leq\frac{s^m(\eta^m-1)}{2^n}.
\end{align*}
Finally, do the summation as in the $m=1$ case.
A: Here is a slight simplification of @triple_sec's answer: Fix $n > m$. Then there exists $t_0 > 0$ such that
$$ P(X > \eta t) \leq \eta^{-n} P(X > t) \quad \text{for all} \quad t \geq t_0. $$
So, for any $t \geq t_0$,
\begin{align*}
P(X > t)
&= P(X > t_0 \eta^{\log_{\eta}(t/t_0)}) \\
&\leq P(X > t_0 \eta^{\lfloor \log_{\eta}(t/t_0) \rfloor}) \\
&\leq \eta^{-n \lfloor \log_{\eta}(t/t_0) \rfloor} P(X > t_0) \\
&\leq C t^{-n}
\end{align*}
for some constant $C > 0$. (We may pick $C = (\eta t_0)^n P(X > t_0)$, although its value is not important.) Then
\begin{align*}
E[X^m]
&= E\biggl[ \int_{0}^{\infty} mt^{m-1} \mathbf{1}_{\{t < X\}} \, \mathrm{d}t \biggr] \\
&= \int_{0}^{\infty} mt^{m-1} P(t < X) \, \mathrm{d}t \tag{Fubini} \\
&\leq \int_{0}^{t_0} mt^{m-1} \, \mathrm{d}t + \int_{t_0}^{\infty} Cm t^{-(n-m+1)} \, \mathrm{d}t \\
&= t_0^m + \frac{Cm}{n-m}\frac{1}{t_0^{n-m}},
\end{align*}
which is finite. $\square$
