Let $x:=\limsup \frac{a_1+a_2+...+a_{n+1}}{a_n}$ . Let $\epsilon\gt 0$ be given.
If $x$ is infinite then there is nothing to prove so let's consider the case when $x$ is finite.
There exists a $K$ such that for large $n\ge K$ we have $\frac{a_1+a_2+...+a_{n+1}}{a_n}\lt x+\epsilon\implies \frac{a_n}{a_1+a_2+...+a_{n+1}}\gt \frac 1{ x+\epsilon}\overbrace{\implies}^{*} \frac {a_n}{a_{n+1}}\gt \frac 1{x+\epsilon}\implies \frac{a_{n+1}}{a_n}\lt x+\epsilon \implies \limsup \frac{a_{n+1}}{a_n}=x.$
Infact at $(*)$, it can also be concluded that $\frac{a_n}{a_r}\gt \frac 1{x+\epsilon}$ for any $r\le n$. That is, $\limsup \frac{a_{n+1}}{a_r}=x$ for every $r\le n$.
I don't understand how to proceed further. Any help on this is much appreciated. Thanks.
Edit: I am using the following definition of $\limsup (x_n)=:y$
- If sequence $(x_n)$ is bounded from above then $y$ is defined to be a number which has the property: For every $\epsilon \gt 0$, there exists $N\in \mathbb N$ such that $n\ge N\implies x_n\lt y+\epsilon$ and that for infinitely many $n$, the inequality $y-\epsilon\lt x_n$ holds.
- If sequence $(x_n)$ is unbounded from above, then $y:=+\infty$.
Or the following definition can also be used:
$\limsup x_n=\lim_{n\to \infty} \sup_{m\ge n} x_m$