How can I show this sequence $u_n$ is divergent: $u_n=\exp( n\log n-(n+\epsilon)\log(n+\epsilon))$

Question

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 $$u_n=e^{nlog(n)-(n+t)log(n+t)}=e^{n\log(n)-n\log(n+t)-t\log(n+t)}\sim ?$$

Answer 1

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.

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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 $0$.

Answer 2

\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 $+\infty$.

Answer 3

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.