Proving a lower bound on the limit superior of a sequence. Prove that for every positive sequence {$a_{n}$}, 
$$\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}\geq 4$$
Also find the sequences {$a_{n}$} for which 4 is attained.
Attempted Solution: 
At the moment, I just have the following clues:  
1.$$b_{n}:=\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}, $$$$c_{n}:=\sup \left\{b_{m}\mid m\geq n\right\} , $$$$\rightarrow\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\lim_{n \to \infty}c_{n}$$
2.$$b_{n}>1\rightarrow c_{n}>1\rightarrow\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}>1$$
3. Subproblem: Is it true that for positive sequences {$a_{n}$},
$$\lim_{n \to \infty}a_{n}= \infty\to \lim_{n \to \infty}  \frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\infty$$
If yes, then perhaps $\varlimsup_{n \to \infty}\frac{\sum_{i=1}^{n+1}a_{i}}{a_{n}}=\infty$and the result is proven for all positive unbounded sequences {$a_{n}$}.
Kindly provide me hints so that I can progress further.
 A: Suppose that for some $N$, if $n\ge N$
$$
s_{n+1}=\sum_{k=1}^{n+1}a_k\le ca_n=c(s_n-s_{n-1})\tag{1}
$$
Let
$$
f(x)=\sum\limits_{k=N}^\infty s_kx^{k-N}\tag{2}
$$
Then, for $x\ge0$,
$$
\begin{align}
\sum_{k=N}^\infty s_{k+1}x^{k-N}
&\le\sum_{k=N}^\infty cs_kx^{k-N}-\sum_{k=N}^\infty cs_{k-1}x^{k-N}\\[6pt]
\frac{f(x)-s_N}{x}&\le cf(x)-cxf(x)-cs_{N-1}\\[12pt]
(cx^2-cx+1)f(x)&\le s_N-cs_{N-1}x\tag{3}
\end{align}
$$
If $c\lt4$, then $f(x)\le\frac{4s_N}{4-c}$, but since $s_k$ is an increasing sequence, $f(x)$ cannot be bounded. Thus, $c\ge4$; in other words
$$
\limsup_{n\to\infty}\frac{\sum\limits_{k=1}^{n+1}a_k}{a_n}\ge4\tag{4}
$$
Setting $c=4$ in inequality $(3)$ suggests we consider
$$
\begin{align}
f(x)
&=\frac1{(1-2x)^2}\\
&=\sum_{n=0}^\infty(n+1)(2x)^n\tag{5}
\end{align}
$$
This leads us to notice that for sequences where $\frac{a_{n+1}}{a_n}\to2$, we get
$$
\sum_{k=1}^{n+1}a_k\sim4a_n\tag{6}
$$
A: Proof: Assume first that 

$$\varlimsup_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}=+\infty$$
  then we have
  $$\varlimsup_{n \to \infty}\dfrac{a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}}{a_{n}}=+\infty$$
  Now let
  $$\varlimsup_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}=a<+\infty$$
  Then, given $\varepsilon>0$, there exists $k$ such that
  $$\dfrac{a_{n+1}}{a_{n}}<a+\varepsilon,n\ge k$$
  In other words,
  $$\dfrac{a_{n}}{a_{n+1}}>\dfrac{1}{a+\varepsilon},n\ge k$$
  Hence, for sufficiently large $n$, we have
  \begin{align*}
b_{n}&=\dfrac{a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}}{a_{n}}\ge\dfrac{a_{k}+\cdots+a_{n}+a_{n+1}}{a_{n}}\\
&=\dfrac{a_{k}}{a_{k+1}}\cdot\cdots\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{k+1}}{a_{k+2}}\cdots\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}\\
&+\cdots+\dfrac{a_{n-2}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n-1}}{a_{n}}+1+\dfrac{a_{n+1}}{a_{n}}\\
&\ge\left(\dfrac{1}{a+\varepsilon}\right)^{n-k}+\left(\dfrac{1}{a+\varepsilon}\right)^{n-k-1}+\cdots+\dfrac{1}{a+\varepsilon}+1+\dfrac{a_{n+1}}{a_{n}}
\end{align*}
  if $0<a<1$, then the above inequality have
  $$\varlimsup_{n \to \infty}b_{n}=+\infty$$
  on the other hand,if $a>1$, then we have
  $$\varlimsup_{n \to \infty}b_{n}=a+\lim_{n\to\infty}\dfrac{1-\left(\dfrac{1}{a+\varepsilon}\right)^{n-k+1}}{1-\dfrac{1}{a+\varepsilon}}=a+\dfrac{a+\varepsilon}{a+\varepsilon-1}$$
  In case $a=1$($\varepsilon>0$ can be arbitrary) we get
  $$\varlimsup_{n \to \infty}b_{n}=+\infty$$
  if
  $a>1$, then we have
  $$\varlimsup_{n \to \infty}b_{n}\ge 1+a+\dfrac{1}{a-1}=2+(a-1)+\dfrac{1}{a-1}\ge 4$$
  $4$ is an optimal estimate because it is attained for the sequence $$a_{n}=2^n,n\in N$$

