Prove $\lim_{n\to \infty}\left(1-\frac{\ln n}{n}-\sqrt[n]{a_n}\right)\frac{n}{\ln \ln n}=\beta$ For a positive series $\sum_{n=1}^\infty a_n$,if
\begin{align*}
\lim_{n\to \infty}\left(1-\frac1n-\frac{a_{n+1}}{a_n}\right)\cdot n\ln n=\beta
\end{align*}
then
\begin{align*}
\lim_{n\to \infty}\left(1-\frac{\ln n}{n}-\sqrt[n]{a_n}\right)\frac{n}{\ln \ln n}=\beta
\end{align*}
The condition in this question is something like Bertrand's test(see https://math.feld.cvut.cz/mt/txte/2/txe3ea2d.htm),but I have no idea how to prove it.
 A: *

*Step 1

You can assume that $a_n=\frac{1}{n\ln ^p n}$ and you will find that $\beta =p$ by calculation. Due to this reason, in the context of my proof, $\beta$ and p mean the same thing.
\begin{align}
&(1-\frac{1}{n}-\frac{n\ln^pn}{(n+1)\ln^p(n+1)})n\ln n\\=
&((1-\frac{n\ln^pn}{(n+1)\ln^p(n+1)})n-1)\ln n \\=
&(n\frac{(n+1)\ln^p(n+1)-n\ln^pn}{(n+1)\ln^p(n+1)}-1)\ln n
\end{align}
Consider Taylor series of function $f(x)=\frac{1}{x\ln^p x}$, one get
$$f(x)=f(x+1)+(x-x-1)f'(x+1)+(x-x-1)^2f''(\xi)/2 ,\ \  \xi\in(x,x+1)$$
Hence
\begin{align}
&(n\frac{(n+1)\ln^p(n+1)-n\ln^pn}{(n+1)\ln^p(n+1)}-1)\ln n\\=
&(n\frac{\ln^p(n+1)+p\ln^{p-1}(n+1)+f''(\xi)/2}{(n+1)\ln^p(n+1)}-1)\ln n
\end{align}
Let $n \to \infty$, one can get that $$(n\frac{\ln^p(n+1)+p\ln^{p-1}(n+1)+f''(\xi)/2}{(n+1)\ln^p(n+1)}-1)\ln n=p$$
For equation
$$\lim_{n\to \infty}(1-\frac{\ln n}{n}-\sqrt[n]{a_n})\frac{n}{\ln \ln n}=\beta,$$
the proof is similar. Notice that $\sqrt[n]{a_n}=e^{\frac{1}{n}\ln a_n}$ and consider the Taylor series of it.

*

*Step 2

For any positive series $\{a_n\}$ satisfied first equation, one can check that for any $\epsilon >0,p>0$, there exsits $N$ large enough such that for any $n>N$, the following equation hold.
$$\frac{\frac{1}{(n+1)\ln ^{p-\epsilon} (n+1)}}{\frac{1}{n\ln ^{p-\epsilon} n}} > \frac{a_{n+1}}{a_n} > \frac{\frac{1}{(n+1)\ln ^{p+\epsilon} (n+1)}}{\frac{1}{n\ln ^{p+\epsilon} n}} \tag{1}$$
If this affirmance does't hold. Say, for any $N>0$, there always exists $n>N$ such that
$$\frac{\frac{1}{(n+1)\ln ^{p-\epsilon} (n+1)}}{\frac{1}{n\ln ^{p-\epsilon} n}} < \frac{a_{n+1}}{a_n}$$
Then for such $n$,
$$(1-\frac{1}{n}-\frac{n\ln^{p-\epsilon}n}{(n+1)\ln^{p-\epsilon}(n+1)})n\ln n > (1-\frac{1}{n}-\frac{a_{n+1}}{a_n})n \ln n$$
Consider the liminf of both sides of the inequality, one gets
\begin{align}
p-\epsilon \geq &\liminf (1-\frac{1}{n}-\frac{a_{n+1}}{a_n})n \ln n\\
&=\lim_{n\to \infty} (1-\frac{1}{n}-\frac{a_{n+1}}{a_n})n \ln n\\
&=p
\end{align}
Which makes an contradiction

*

*Step 3

Utilize the inequality (1), you can find that $\sqrt[n]{a_n}$ can be controlled by $\sqrt[n]{n \ln^{p\pm \epsilon}n}$. Specificlly
\begin{align}
\frac{a_{n+k}}{a_n}
=&\frac{a_{n+k}}{a_{n+k-1}}\cdot \frac{a_{n+k-1}}{a_{n+k-2}}\dots \frac{a_{n+1}}{a_{n}}\\
>&\frac{\frac{1}{(n+k)\ln ^{p+\epsilon} (n+k)}}{\frac{1}{n\ln ^{p+\epsilon} n}}
\end{align}
Fix $n$ and let $k$ become big enough, one gets
$$\sqrt[n+k]{a_{n+k}}>\sqrt[n+k]{n\ln^{p+\epsilon} n \cdot a_n} \cdot \sqrt[n+k]{\frac{1}{(n+k)\ln^{p+\epsilon} (n+k)}}$$
Therefor,for $k$ is large enough, we get
$$\sqrt[n+k]{a_{n+k}}>\sqrt[n+k]{\frac{1}{(n+k)\ln^{p+3\epsilon/2} (n+k)}}$$
Hence for $n$ is large enough, we have
$$\sqrt[n]{\frac{1}{n\ln^{p+\epsilon}n}}<\sqrt[n]{a_n}<\sqrt[n]{\frac{1}{n\ln^{p-\epsilon}n}}$$
Use this inequality combined with the conclusion we get in step 1, we can get that
$$p-\epsilon <(1-\frac{\ln n}{n}-\sqrt[n]{a_n})\frac{n}{\ln \ln n} <p +\epsilon$$
