Let $a_n,b_n\subset \mathbb R$ be two sequences, where $\lim a_n=a\in \mathbb R$ a and $\lim b_n$ doesn't exist. What can we say about $$\lim(a_n+b_n)$$ and $$\lim(a_n\cdot b_n)$$ I think that $\lim(a_n+b_n)$ doesn't exist and $\lim(a_n\cdot b_n)$ has limit (0) only if $\lim a_n = 0$, but I don't know how to prove it. (how to get those solutions)
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$\begingroup$ For the first one, if the limit existed, what could you say about $\lim\left((a_n+b_n)_{n\in \mathbb N}\right)-\lim\left((a_n)_{n\in \mathbb N}\right)$? The second one, for $a\neq 0$ is similar. But if $a=0$ your remark is false if you set $a_n=\dfrac 1 n$ and $b_n=n$, for all $n\in \mathbb N$. $\endgroup$– Git GudNov 10, 2014 at 1:35
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$\begingroup$ You can't set $b_n=n$ because then it has limit in $\infty$. But you can set $b_n=(-1)^n\cdot n$. (of course what I thought firstly is bad) but now I'd like to know how to distinguish whether it has limit or no for $a=0$. $\endgroup$– BreldorNov 10, 2014 at 3:02
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1 Answer
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Suppose $a_n \to a$. Then if $a_n+b_n \to c$, you would have $b_n = (a_n+b_n)-a_n \to c-a$, which would be a contradiction.
Suppose $a \neq 0$, then if $a_n b_n \to c$, you would have $b_n = (a_n b_n) {1 \over a_n} \to c {1 \over a}$, which would be a contradiction.
For the last, take $b_n = (-1)^n n$, $a_n = {1 \over n}$. Then $a_n b_n$ has no limit.
answered Nov 10, 2014 at 1:35
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$\begingroup$ Why do we need that third case? $b_{n}$ doesn't have limit so it can't be $=n$. $\endgroup$– BreldorNov 10, 2014 at 1:47
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$\begingroup$ I'd like to know how to distinguish whether it has limit or no for $a=0$. Any hint? $\endgroup$– BreldorNov 10, 2014 at 3:03
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$\begingroup$ Well, $\infty$ isn’t usually considered a bona fide limit unless you are working in the extended reals. However, I have modified the answer so that $b_n$ doesn't have a limit (unless one considers $\infty$ as in the Riemann sphere...). $\endgroup$ Nov 10, 2014 at 3:17