Suppose $ \lim \left( a_{n}+a_{n+1} \right)=0 $. Show that $ \lim a_{n}=0 $ or $ 0 < \limsup a_{n} $. I tried separating into cases, where $ a_{n} $ converges (thus is bounded), is unbounded (thus diverges) and is bounded and also diverges. I also tried proving by way of contradiction.
I am having trouble still with showing that if $ \lim a_{n}\neq0 $ and $ \limsup a_{n}\le0 $ and $ a_{n} $ is bounded and diverges. then $ \lim \left( a_{n}+a_{n+1} \right) \neq0  $.
 A: $\limsup a_n$ is the sup of all limits of subsequences of $a_n$, which sup may be $-\infty$ or $+\infty$ (Baby Rudin 3.16).  
Suppose there exists an unbounded subsequence.  Then we can find a subsequence going either to $-\infty$ or $+\infty$.  If it's $+\infty$ we're done.  If $a_{n_i}$ goes to $-\infty$, then because $\lim (a_n+a_{n+1}) = 0$, for large $n$ we have $a_{n_i+1}\geq -a_{n_i}-1$, and so $a_{n_i+1}$ goes to $+\infty$ and we're done. 
Suppose now that all subsequences are bounded, say lying in $[-A,A]$ for some finite positive $A$.  There have to be convergent subsequences (Baby Rudin 2.41).  If a subsequence converges to a positive value we're done.  Suppose that a subsequence $a_{n_i}$ converges to a value $a < 0$:  then because $\lim (a_n+a_{n+1}) = 0$ we must have $a_{n_i+1}$ converging to $-a > 0$ and we're done.  
We are left with the possibility that the only limit of subsequences is $0$.  In that case, there can be no point of accumulation in $[-A,-\epsilon]$ or $[\epsilon,A]$ for any $\epsilon>0$, and so all but a finite number of $a_n$'s lie outside those intervals, and $a_n\to 0$.
A: I will use the following auxiliary results:
Lemma 1: For any sequences $(a_n),(b_n)$ we have
$$\liminf a_n+\liminf b_n\leq\liminf(a_n+b_n)\leq\limsup(a_n+b_n)\leq\limsup a_n+\limsup b_n\,.$$
Lemma 2: For any sequence $(a_n)$ we have
$$\liminf(-a_n)=-\limsup a_n\,.$$
Assuming these results, take $b_n=a_{n+1}$, in which case we have $\liminf a_n=\liminf b_n$ and $\limsup a_n=\limsup b_n$. The hypothesis, together with last inequality in Lemma 1, imply that $\limsup a_n\geq0$. Finally, if $\limsup a_n=0$ then $\liminf(-a_n)=-\limsup a_n=0$, so $\liminf b_n=\liminf\bigl[(a_n+b_n)+(-a_n)]\geq\liminf(a_n+b_n)+\liminf(-a_n)=0$. Thus $0\leq\liminf b_n\leq\limsup b_n=0$, which yields $\lim b_n=0$.
