Show the sequence $\{a_{n}\}_{n=1}^{\infty}$ is unbounded 
Let $a_{n}>0 ,(n=1,2,\cdots)$ and   $$\lim_{n\rightarrow\infty}\frac{a_{_{n}}}{a_{_{n+1}}+a_{_{n+2}}}=0.$$Show the sequence $\{a_{n}\}_{n=1}^{\infty}$ is unbounded.

The following is my thoughts.But I am not sure my answer is right,I need someone to check it.Even though it's correct, this answer is complicated and cumbersome.Do you have some concise ways by using reduction to absurdity? 

Suppose the sequence $\{a_{n}\}$ is bounded,then exists $M(>0)$ such that $0<a_{n}\leq M,(n=1,2,\cdots).$
(1). 
$$0<\frac{a_{n}}{2M}\leq\frac{a_{_{n}}}{a_{_{n+1}}+a_{_{n+2}}},(n=1,2,\cdots)\Rightarrow\lim_{n\rightarrow\infty}a_{n}=0 $$
(2). 
$$\frac{a_{n}+a_{n+1}}{a_{n+1}+a_{n+2}}=A(n)+B(n),A(n)=\frac{a_{n}}{a_{n+1}+a_{n+2}}, B(n)=\frac{a_{n+1}}{a_{n+1}+a_{n+2}}(B(n)\in (0,1],n=1,2,\cdots)\Rightarrow$$ $$0\leq\varlimsup_{n\rightarrow\infty} \frac{a_{n}}{a_{n+2}+a_{n+3}}=\varlimsup_{n\rightarrow\infty} \frac{a_{n}}{a_{n+1}+a_{n+2}}\cdot \frac{a_{n+1}+a_{n+2}}{a_{n+2}+a_{n+3}}\leq\varlimsup_{n\rightarrow\infty} \frac{a_{n}}{a_{n+1}+a_{n+2}} \cdot\varlimsup_{n\rightarrow\infty} \frac{a_{n}+a_{n+1}}{a_{n+1}+a_{n+2}}=0\Rightarrow \lim_{n\rightarrow\infty}\frac{a_{n}}{a_{n+2}+a_{n+3}}=0$$
(3). Futher,we have  $\forall p\in\mathbb{N},$ 
$$\lim_{n\rightarrow\infty}\frac{a_{_{n}}}{a_{_{n+p}}+a_{_{n+p+1}}}=0$$
(4).
 From (3),Let $$p_{1}=1, \exists N_{1}\in\mathbb {N},\text{such that}\frac{a_{_{n}}}{a_{_{n+1}}+a_{_{n+2}}}<\frac{1}{2}\text{ whenever} \quad n>N_{1};$$
$$p_{2}=2, \exists N_{2}\in\mathbb {N}(N_{2}>N_{1}),\text{such that}\frac{a_{_{n}}}{a_{_{n+2}}+a_{_{n+3}}}<\frac{1}{2}\text{ whenever} \quad n>N_{2};$$ $$\cdots\cdots\cdots\cdots$$$$p_{k}=k, \exists N_{k}\in\mathbb {N}(N_{k}>N_{k-1}>\cdots>N_{1}),\text{such that}\frac{a_{_{n}}}{a_{_{n+k}}+a_{_{n+k+1}}}<\frac{1}{2}\text{ whenever} \quad n>N_{k};$$$$\cdots\cdots\cdots\cdots$$
(5). From (4),there must exists an index $k_{0}\in \mathbb{N}$ and $n_{0}>N_{k_{0}}$ such that $a_{n_{0}}>a_{n_{0}+k_{0}}$ and $a_{n_{0}}>a_{n_{0}+k_{0}+1}.$ In fact, If for all $k\in\mathbb{N}$ and every $n>N_{k},$ we have either $a_{n}\leq a_{n+k}$ or $a_{n}\leq a_{n+k+1},$ then for every $n>N_{k},a_{n}=0$.This is contradicting  $a_{n}>0 (n=1,2,\cdots)!$
(6). From (5),
$$\frac{1}{2}<\frac{a_{n_{0}}}{a_{n_{0}+k_{0}}+a_{n_{0}+k_{0}+1}}<\frac{1}{2}.
\quad \text{It's impossible!}$$

From above all,we can say the sequence $\{a_{n}\}_{n=1}^{\infty}$ is unbounded.
 A: By the limit hypothesis, for $n$ large enough,
\begin{equation}
\frac{a_n}{a_{n+1} + a_{n+2}} < \frac{1}{4}.
\end{equation}
Therefore, either $a_{n+1} > 2 a_n$ or $a_{n+2} > 2 a_n$.  Using that, we can find a subsequence $a_{i_0}, a_{i_1}, \ldots$ such that $a_{i_n} \ge 2^n a_{i_0}$ (by choosing $i_{n+1}$ to be either $i_n + 1$ or $i_n + 2$).  That easily implies that the subsequence is unbounded, and therefore the original sequence is also unbounded.
A: Suppose the sequence $\{a_{n}\}$ is bounded, that is to say,
there exists $M>0$ such that $$0<a_{n}\leq M,\ \ n=1,2,\cdots.$$
So
$$0<\frac{a_{n}}{2M}\leq\frac{a_n}{a_{n+1}+a_{n+2}}\implies \lim_{n\to\infty}a_{n}=0.$$
By
$$\lim_{n\to\infty}\frac{a_n}{a_{n+1}+a_{n+2}}
=\lim_{n\to\infty}a_{n}=0,$$
we know: for $\forall \epsilon>0,\exists N>0$, when $n>N$,
$$a_n<\epsilon, \qquad \color{red}{\frac{a_n}{a_{n+1}+a_{n+2}}<\epsilon}.$$
So for any $k\geq1$,
\begin{align*}
a_{_{N+1}}
&\color{red}{<}(a_{_{N+2}}+a_{_{N+3}})\epsilon\\
&\color{red}{<}(a_{_{N+3}}+2a_{_{N+4}}+a_{_{N+5}})\epsilon^2\\
&\color{red}{<}(a_{_{N+4}}+3a_{_{N+5}}+3a_{_{N+6}}+a_{_{N+7}})\epsilon^3\\
&\color{red}{<}\cdots\\
&\color{red}{<}M\epsilon^k\left(1+\binom{k}{1}+\binom{k}{2}+\cdots+\binom{k}{k}\right)\\
&=M(2\epsilon)^k,
\end{align*}
Take $\epsilon=1/4$, then, for any $k\geq1$, we konw
$$0<a_{_{N+1}}<\frac{M}{2^k},$$
let $k\to\infty,$ we get $a_{_{N+1}}=0$, which is in contradiction with $a_{_{N+1}}>0$.
