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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 $.

Martin Sleziak
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PalmTopTigerMO
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    If $\limsup a_n<0$, there will be a contradiction. If $\limsup a_n=0$, prove $\liminf a_n=0$ by contradiction. – Danielsen Dec 09 '15 at 02:10

2 Answers2

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$\limsup a_n$ 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 $a_n\to 0$.

ForgotALot
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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 $\lim b_n=0$.

Matemáticos Chibchas
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  • Some posts related to Lemma 1: http://math.stackexchange.com/questions/408135/prove-limsup-limits-n-to-infty-a-nb-n-le-limsup-limits-n-to-infty and http://math.stackexchange.com/questions/70478/properties-of-liminf-and-limsup-of-sum-of-sequences (And also questions linked to the first post and the questions linked to the second one.) – Martin Sleziak Dec 09 '15 at 05:10
  • Not all too related to my original post, but my text book gives a condition for the sequences in the above lemmas to be bounded. Does it do this because I'm currently in baby phase of the learning of the subject? – PalmTopTigerMO Dec 10 '15 at 09:56