the convergence of the sequence $x_n = n^7/7^n$ Let $(x_n)_{n\in \mathbb{N}}$ be the sequence 
\begin{align}
x_n = \frac{n^7}{7^n}.
\end{align}
It is expected that $x_n \to 0$ as $n\to +\infty$ since the denominator goes to $+\infty$ faster than the numerator as $n$ gets larger and larger. My question is: How can I prove that $x_n\to 0$ from the definition of the convergence (The $\varepsilon-n_0$ definition). 
Thanks in advance.
 A: Hint By Bernoulli inequality,
$$\left( \frac{n}{n+1} \right)^7=\left( 1-\frac{1}{n+1} \right)^7\geq 1-\frac{7}{n+1}$$
Therefore, for $n \geq 13$ you have
$$\left( \frac{n}{n+1} \right)^7 \geq 1-\frac{7}{14} =\frac{1}{2}$$
Use this to show that for all $n \geq 14$ you have 
$$\frac{x_{n+1}}{x_n} \leq \frac{2}{7} $$
Deduce from here that
$$x_n \leq x_{14} \left( \frac{2}{7} \right)^{n-14} \, \forall n \geq 14$$
Now, use th $\epsilon -N$ definition. If you know logarithms, you can find $N_\epsilon$ by using logarithms. Otherwise, use Bernoully again
$$\left( \frac{7}{2} \right)^{n-14} \geq 1+(n-14) \frac{5}{2} \Rightarrow \\
 0 \leq x_n \leq x_{14} \left( \frac{2}{7} \right)^{n-14} \leq \frac{x_{14}}{ 1+(n-14) \frac{5}{2} }$$
Note If you learned series, a faster solution is the following
$$\sum_n x_n $$
is convergent by the Ratio Test. Therefore, by the divergence test, 
$$\lim_n x_n =0$$
A: More generally,
if
$x_n = \dfrac{n^a}{b^n}$
where $a > 0, b > 1$
then
$\dfrac{x_{n+1}}{x_n}
=\dfrac{\dfrac{(n+1)^a}{b^{n+1}}}{\dfrac{n^a}{b^n}}
=\dfrac{(1+1/n)^a}{b}
$.
Therefore
$\begin{array}\\
\dfrac{x_{n+m}}{x_n}
&=\prod_{k=0}^{m-1} \dfrac{x_{n+k+1}}{x_{n+k}}\\
&=\prod_{k=0}^{m-1} \dfrac{((n+k+1)/(n+k))^a}{b}\\
&=\dfrac{((n+m)/n)^a}{b^m}\\
&=\dfrac{(1+m/n)^a}{b^m}\\
\end{array}
$
so,
since $\ln(1+x) < x$,
$\begin{array}\\
\ln(\dfrac{x_{n+m}}{x_n})
&=\ln(\dfrac{(1+m/n)^a}{b^m})\\
&=a\ln(1+m/n)-m\ln(b)\\
&\lt am/n-m\ln(b)\\
&= m(a/n-\ln(b))\\
\end{array}
$
If $a/n < \ln(b)/2$,
or $n > 2a/\ln(b)$
then
$\ln(\dfrac{x_{n+m}}{x_n})
\lt -m\ln(b)/2
$
so that,
if $\ln(x_n)/m-\ln(b)/2 < -\ln(b)/4$
or
$m > 2\ln(x_n)/\ln(b)
$ 
$\begin{array}\\
\ln(x_{n+m})
&\lt \ln(x_n)-m\ln(b)/2\\
&= m( \ln(x_n)/m-\ln(b)/2)\\
&< -m\ln(b)/4\\
&=\ln(b^{-m/4})
\text{so}\\
x_{n+m}
&\lt \dfrac1{b^{m/4}}\\
&\to 0\\
\end{array}
$
as $m \to \infty$.
