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In course of solving the problem How to show that $\lim \sup a_nb_n=ab$ I feel that I've probably made some mistake in my solution for I didn't use the fact that $a_n>0$ $\forall$ $n\geq1.$

Please help me to find out the mistake I've made:

  • $\exists$ a subsequence $\{a_{r_n}\}$ of $\{a_n\}$ such that $a_{r_n}\to a.$ Now $a_{r_n}\to a, b_{r_n}\to b\implies a_{r_n}b_{r_n}\to ab\implies ab$ is a subsequencial limit of {$a_nb_n$}. If possible let $\exists$ a subsequence $\{a_{p_n}b_{p_n}\}$ of $\{a_nb_n\}$ such that $a_{p_n}b_{p_n}\to m>ab.$ Since $b_n,b>0$ so $b_n^{-1}\to b^{-1}>0$ whence $a_{r_n}\to mb^{-1}>a,$ a contradiction to $\lim \sup a_n=a.$ Hence the result follows.

Thanks for voting up the question. But that didn't actually eliminate my confusion. I'm looking for some concrete comments and opinions.

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What are the definition of $a$ and $b$? – Wayson Kong Jan 10 '13 at 13:31
$a=\lim\sup a_n,b=\lim b_n.$ Follow the link given in the question. – Sugata Adhya Jan 10 '13 at 13:35
why not try using the definition of Cauchy sequence? – Santosh Linkha Jan 10 '13 at 13:43
@experimentX: Is the definition I used wrong? – Sugata Adhya Jan 10 '13 at 13:47
In the original problem… the existence of $\lim_{n\to\infty} b_n$ was not assumed. Your assumption is much stronger, since without it you cannot conclude the convergence of $a_{p_n}$ from the convergence of $a_{p_n}b_{p_n}$. – Arin Chaudhuri Jan 11 '13 at 3:43
up vote 0 down vote accepted

You do need some extra assumption. Consider the following example (where $a$ is a finite real):

$a_n=a$ if $n$ is odd, $a_n=-n$ if $n$ is even,

$b_n=-1.$ (so in notation, $\lim b_n=b=-1$)

Then lim sup of $a_n$ is $a$, and lim $b_n$ is b, yet we have

$a_nb_n=-a$ if $n$ is odd, $a_nb_n=n$ if $n$ is even.

So in this example the limsup of $a_nb_n$ is $+\infty$, whereas $ab$ is not.

EDIT: I noted that in your proof you needed to use $b>0$ but that assumption should be stated in the problem.

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Your proof is correct. You proved that:

If $\lim\sup a_n=a, \ \lim b_n=b>0$ then $\lim\sup a_nb_n=ab$.

which is true.

The assumptions $a_n>0$ and $b_n>0, \ \forall n\geq1$ are required for this proposition:

If $\lim\sup a_n=a, \ \lim\sup b_n=b$ then $\lim\sup a_nb_n=ab$.

Also see the last corollary here.

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My solution :

Case-1 : Let $lim \sup u_n = l$ and $lim \sup v_n = m$.First of all let us assume that at least one of $l$ or $m$ be $0$.Without loss of generality we assume that $l = 0$ then it is obvious that $\{u_n\}$ should converge to $0$.Since $\{v_n\}$ is bounded so the sequence $\{u_n v_n\}$ should converge to $0$ and from here it can be easily verified that the given condition hold.

Case-2 : Let none of $l$ or $m$ be $0$.Let us choose $0 < \epsilon < 8lm$.Then for this chosen $\epsilon > 0$ $ \exists p, q \in \mathbb {N}$ such that $\forall_n \geq p$ and $\forall_n \geq q$, $u_n < l + \frac \epsilon {4m}$ and $v_n < m + \frac \epsilon {4l}$ respectively.Let $k = max \{p , q\}$.Then $u_n v_n < lm + \frac {\epsilon( 1 + 1)} 4 + (\epsilon)^2/16lm < lm + \epsilon$, $\forall_n \geq k$ and hence the result follows since $\epsilon$ can be taken arbitrarily small in the interval $(0,8lm)$.Look that $(\epsilon)^2/{16lm} < \frac \epsilon 2$.

Hence, $lim \sup u_n v_n \leq lm$.

In the entire process I have assumed that $u_n>0$ and $v_n>0$, $\forall_n \in \mathbb {N}$ .So your statement is in general not true.

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