How can I show this sequence $u_n$ is divergent: $u_n=\exp( n\log n-(n+\epsilon)\log(n+\epsilon))$ How can I show this sequence $u_n$ is divergent:
$$u_n=\exp( n\log n-(n+\epsilon)\log(n+\epsilon))\quad n\in \mathbb{N}^*;\quad \epsilon \in (0,1)$$
My attempts:
\begin{align*}
u_n&=\exp( n\log n-(n+\epsilon)\log(n+\epsilon))\\
u_n&=\exp( n(\log n-\log(n+\epsilon))-\epsilon \log(n+\epsilon))\\
u_n&=-\exp( n(-\log n+\log(n+\epsilon))+\epsilon \log(n+\epsilon))\\
&\leq -\epsilon \log(n+\epsilon)\\
\end{align*}
note that
$-\epsilon \log(n+\epsilon)\to -\infty$
but i'm stuck here
Edit
$$u_n =e^{n\log(n)-(n+e)\log(n + e)}$$ 
$$\log(n+t)=\log(n(\frac{t}{n}+1))=\log(n) + \frac{t}{n} +o(\frac{t}{n})$$
can we say that
$$n\log(n+t)=n\log(n) + t +o(t)$$
then
$$u_n=e^{nlog(n)-(n+t)log(n+t)}=e^{n\log(n)-n\log(n+t)-t\log(n+t)}\sim ?$$
 A: When $n>1$, both $(n+\epsilon)$ and $\log(n+\epsilon)$ are positive increasing functions in $\epsilon$, i.e.
$(n+\epsilon)\log(n+\epsilon)>n\log n$ 
and hence 
$n\log n-(n+\epsilon)\log(n+\epsilon)<0.$ 
Taking $\exp$ of something negative results in something smaller than $1$, and so the expression doesn't diverge after all.

In more details, let's first write
${\mathrm e}^{n\log n-(n+\epsilon)\log(n+\epsilon)}=\dfrac{1}{{\mathrm e}^{n\log(n+\epsilon)+\epsilon\log(n+\epsilon)-n\log n}}=\dfrac{1}{{\mathrm e}^{n\log(n+\epsilon)-n\log n}}\dfrac{1}{(n+\epsilon)^\epsilon}.$
In ${\mathrm e}^{n\log(n+\epsilon)-n\log n}$, we can get rid of $n\log(n)$ by observing that
$n\log(n+\epsilon)$
$=n\log(\frac{n+\epsilon}{n}\cdot n)$
$=n \log(1+\frac{\epsilon}{n})+n\log(n)$
$=\log(\left(1+\frac{\epsilon}{n}\right)^n)+n\log(n)$
and so the expression becomes
$\dfrac{1}{\left(1+\frac{\epsilon}{n}\right)^n}\dfrac{1}{(n+\epsilon)^\epsilon}.$
The limit of $\left(1+\frac{\epsilon}{n}\right)^n$ is the constant ${\mathrm e}^\epsilon$ but $(n+\epsilon)^\epsilon$ blows up and hence the expression becomes $0$.
A: \begin{align*}
(n+\varepsilon)\log (n+\varepsilon)=&(n+\varepsilon)\log\left(n\left(1+\frac{\varepsilon}{n}\right)\right)\\
=&(n+\varepsilon)\log n+(n+\varepsilon)\log\left(1+\frac{\varepsilon}{n}\right)\\
=&(n+\varepsilon)\log n+\underbrace{\frac{(n+\varepsilon)}{n}\log\left(1+\frac{\varepsilon}{n}\right)^n}_{=:A_{\varepsilon}(n)}
\end{align*}
Hence 
\begin{align*}
n\log n-(n+\varepsilon)\log (n+\varepsilon)
=&n\log n-(n+\varepsilon)\log n-A_{\varepsilon}(n)\\
=&-\varepsilon\log n-A_{\varepsilon}(n)\\
=&\log n^{-\varepsilon}-A_{\varepsilon}(n)
\end{align*}
Thus
\begin{align*}
u_n
=&\exp\left({\log n^{-\varepsilon}-A_{\varepsilon}(n)}\right)\\
=&\exp(\log n^{-\varepsilon})\exp(-A_{\varepsilon}(n))\\
=&\frac1{n^{\varepsilon}e^{A_{\varepsilon}(x)}}
\end{align*}
Ok: now
$$
\lim_{n\to+\infty}A_{\varepsilon}(n)=\varepsilon
$$
and this holds for all $\varepsilon>0$ fixed.
Moreover $n^{\varepsilon}$ goes to infinity, for every $\varepsilon>0$ fixed.
Thus $u_n\to 0$ as $n$ approaches to $+\infty$.
A: Note that
$$
e^{(n+\varepsilon)\log(n+\varepsilon)}=(n+\varepsilon)^{n+\varepsilon}=n^{n+\varepsilon}\left(1+\frac{\varepsilon}{n}\right)^{n}\left(1+\frac{\varepsilon}{n}\right)^{\varepsilon},
$$
so
$$
e^{n\log n - (n+\varepsilon)\log(n+\varepsilon)}=\frac{n^n}{n^{n+\varepsilon}}\left(1+\frac{\varepsilon}{n}\right)^{-n}\left(1+\frac{\varepsilon}{n}\right)^{-\varepsilon}=n^{-\varepsilon}\left(1+\frac{\varepsilon}{n}\right)^{-n}\left(1+\frac{\varepsilon}{n}\right)^{-\varepsilon}.
$$
The middle term converges to $e^{-\varepsilon}$ and the right-hand term converges to $1$ as $n\rightarrow\infty$.  So altogether you have
$$
e^{n\log n - (n+\varepsilon)\log(n+\varepsilon)} \sim (e n)^{-\varepsilon}\rightarrow 0,
$$
using the fact that $\varepsilon > 0$ only in the final step.
