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EDIT: Apparently I translated the terminology wrong, but partial limit means limit of a subsequence.

If $a_{n}$ has 'a' partial limits and $b_{n}$ has 'b', can $a_{n}+b_{n}$ have more than $ab$ partial limits? ('infinity' also counting as a limit)

This is a homework question. I'm thinking the statement is false, but I have no idea where to begin! Could someone give me a hint?

Thanks!

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    "partial limit"?2011-11-17
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    Maybe I translated it wrong? It is a limit of a sub-sequence of $a_{n}$, e.g. '1' and '-1' are partial limits of 1,-1,1,-1,...2011-11-17

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I take it by "partial limit" you mean limit of a subsequence. It is possible for $a_n$ and $b_n$ to have zero partial limits each, while $a_n+b_n$ has a limit. E.g., $a_n=n$, $b_n=-n$.

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    I believe the question refers to limits in the broad sense, that is infinity is a valid limit... I've edited it to say as much!2011-11-17
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    Change $a_n$ to two different sequences which go to infty... $a_{2n}=2n$ and $a_{2n+1}=2n-2$ for example... $b_n=-n$ is unchanged....2011-11-17
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    I'm not sure but the answer might change if your sequences are bounded .....2011-11-17
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    N.S.: In your example $a_{n}$ has one limit (infinity), $b_{n}$ has one limit (negative infinity), and $a_{n}+b_{n}$ has exactly $1*1=1$ limits (infinity), so I suspect this doesn't work.2011-11-17
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    @rowe $a_n+b_n$ has 2 limits ;) Keep in mind that you don't add $a_{2n}$ to $b_n$, you add it to $b_{2n}$ ;)2011-11-17
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    In my example $a_{2n}+b_{2n}=0$ and $a_{2n+1}+b_{2n+1}=-3$....2011-11-17
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    Ooooooh. I'm so stupid sometimes. Thanks! Post this as an answer so I can accept?2011-11-17
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    Don't worry, we all make such small mistakes sometimes ;) And gery gave me the idea, I just changed his answer a little so you should probably accept his answer ;)2011-11-17
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    Okay then. I'm not the best at calculus but how I missed something so simple is beyond me! Thanks everyone.2011-11-17
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    Interestingly enough, it does indeed work for a bounded sequence.2011-11-17
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    @Phira, maybe you could post a proof of that.2011-11-17
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The claim is true if the sequences involved are bounded.

Let $a_{n_k}+b_{n_k}$ be a convergent subsequence of the sum sequence.

Since the sequence $a_{n_k}$ is bounded, it has a converging subsequence $a_{n_{k_i}}$ that converges to some limit point of $a_n$.

This means that $b_{n_{k_i}}$ is the difference of two converging sequences, so it converges as well and its limit is a limit point of $b_n$.

Therefore, the limit of the convergent sequence $a_{n_k}+b_{n_k}$ which is of course the limit of its subsequence is the sum of a limit point of $a_n$ and a limit point of $b_n$.

There are only $a\cdot b$ such sums, therefore the total number is at most $a\cdot b$.