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$a_{n} , b_{n}$ are both bounded series, so both of them have a converged subsequence. I'll define: $(a_{n_k}), (b_{n_k})$ as such: $\lim(a_{n_k}) = \liminf(a_{n}), \lim(b_{n_k}) = \limsup(b_{n})$

Now I can can claim according to the definition of the limsup: $\limsup (a_{n}+b_{n})\ge \lim(a_{n_k} + b_{n_k}) = \lim(a_{n_k}) + \lim(b_{n_k}) = \liminf(a_n) + \limsup(b_{n})$

Martin Sleziak
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    This does not work because there is no reason why a single subsequence $(n_k)$ could provide $\liminf a_n$ and $\limsup b_n$ simultaneously (for a counterexample, consider $a_n=b_n=(-1)^n$). – Did Nov 26 '16 at 23:34
  • By blocked series, I imagine you mean bounded sequence (regardless, Did's comment holds). Also $y_n$ is probably a typo in the title. – parsiad Nov 26 '16 at 23:35
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    Understood the problem. Could you help me and suggest a valid solution? I saw this question already asked here, yet the answers were not informative enough for me to understand –  Nov 26 '16 at 23:39
  • Perhaps it is worth linking to other posts about the same question, for example here or here. – Martin Sleziak Nov 27 '16 at 01:34

2 Answers2

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Note that \begin{equation*} \liminf a_n = -\limsup -a_n. \end{equation*} Then use the fact that \begin{equation*} \limsup c_n +d_n \leq \limsup c_n +\limsup d_n \end{equation*}
Specifically, let $c_n=b_n+a_n$ and $d_n=-a_n$. Then \begin{equation*} \limsup b_n = \limsup (b_n + a_n -a_n) \leq \limsup (b_n+a_n) + \limsup -a_n. \end{equation*} Rearranging terms, we have \begin{equation*} \limsup b_n - \limsup -a_n \leq \limsup (b_n+a_n) \end{equation*} or \begin{equation*} \limsup b_n + \liminf a_n \leq \limsup (b_n+a_n). \end{equation*}

proof of claim: Let $\{c_{n_k}+d_{n_k}\}$ be a subsequence converging to $\limsup_n c_n+d_n$. Now, $\limsup_k c_{n_k} \leq \limsup_n c_n $ and $\limsup_k d_{n_k} \leq \limsup_n d_n $ since $\{n_k\} \subset \mathbb{N}$.

Dunham
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Consider $\{n_k\}$ s.t $(a_{n_k})$ goes to $\limsup (a_n)$. $\{(a+b)_{n_k}\}$ is at least $\limsup(a_n) + \liminf (b_n)$.

Martin Sleziak
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Jacob Wakem
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  • If I understand correctly, you are trying to use the fact that if $x_n$ is convergent, then $\liminf (x_n+y_n)= \lim x_n + \liminf y_n$. (The dual claim for limsup is proved here.) – Martin Sleziak Nov 27 '16 at 03:49
  • @MartinSleziak Nah I'm just trying to use that any "superior subsequence" for a_n has a corresponding subsequence in b_n; thus it is at least an "inferior subsequence". Thus some subsequence is at least the desired value and thus the "superior subsequence" by definition is at least the desired value. (I think I've already answered this question before) – Jacob Wakem Nov 27 '16 at 03:54