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I'm having hard time understanding when I can do the following move: $\lim_{n\to \infty}a_n\cdot b_n = \lim_{n\to \infty}a_n\cdot\lim_{n\to \infty}b_n $

On one hand, my teacher said that I can do that only if prove that both limits of $a_n$ and $b_n$ are exist and finite. On the other hand, When I calculated the limit of the series $a_n=n*cos1/n$ using this site https://www.symbolab.com which is pretty reliable, I saw this steps:

$\lim_{n\to \infty}n\cos(1/n)=\lim_{n\to \infty}n\cdot \lim_{n\to \infty}cos(1/n)="\infty\cdot 1"=\infty$

So how this move is possible if the limit of $a_n=n$ is not finite?

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    $\begingroup$ If $\lim_{n\to +\infty}a_n=L\in\mathbb{R}\setminus\{0\}$, $$ \lim_{n\to +\infty} a_n b_n = L\cdot \lim_{n\to +\infty} b_n $$ where both sides are assumed to be elements of $\mathbb{R}\cup\{-\infty,+\infty\}$. $\endgroup$ – Jack D'Aurizio Dec 24 '17 at 18:06
  • $\begingroup$ If $a_n$ converges to a finite nonzero value and $b_n$ diverges, then $a_nb_n$ will diverge as well. Part of the difficulty is when you have one converging to zero while the other diverges to infinity, the product is unclear without further attention since $0\cdot \infty$ is an indeterminate form. Also unclear is when you have both $a_n$ and $b_n$ diverging (not necessarily diverging to infinity), that doesn't imply convergence or divergence of $a_nb_n$ $\endgroup$ – JMoravitz Dec 24 '17 at 18:06
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    $\begingroup$ As an aside, one could avoid the limit of products is product of limits confusion for your example by noting that for $n\geq 2$ one has $\cos(\frac{1}{n})>0.8$ so for all $n\geq 2$ one has $n\cos(1/n)\geq 0.8 n$ which should clearly diverge to infinity here. $\endgroup$ – JMoravitz Dec 24 '17 at 18:12
  • $\begingroup$ As @JackD'Aurizio points out, the product / quotient rule of limits is valid if limit of one of the operands is non-zero. In case of sum/difference the rule can be applied when one of the operands has a limit. This effectively replaces that operand by its limit and the overall limit becomes dependent on the limit of second operand. $\endgroup$ – Paramanand Singh Dec 25 '17 at 4:54
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You surely can apply $$ \lim_{n\to\infty}a_nb_n= \Bigl(\lim_{n\to\infty}a_n\Bigr) \Bigl(\lim_{n\to\infty}b_n\Bigr) \tag{*} $$ when both limits on the right-hand side exist and are finite.

If one of the limit does not exist, you cannot even write (*) to begin with, so it doesn't make sense to ask whether you can apply it.

However, it's possible to add “rules” when one of those limits is infinity; essentially, if $\lim_{n\to\infty}b_n=c>0$ and $\lim_{n\to\infty}a_n=\infty$, also $\lim_{n\to\infty}a_nb_n=\infty$. Similar (and obvious) rules hold when $c<0$ or the other limit is $-\infty$.

The proof for the extended rule stated above is easy. Since $\lim_{n\to\infty}b_n=c>0$, there exists $N_0$ such that, for $n>N_0$, $$ b_n>\frac{c}{2} $$ Now take $K>0$; since $\lim_{n\to\infty}a_n=\infty$, there exists $N$ such that $N>N_0$ and, for all $n>N$, $a_n>2K/c$. Then, for $n>N$, $$ a_nb_n>\frac{2K}{c}\frac{c}{2}=K $$

In your case, $$ \lim_{n\to\infty}n=\infty \qquad \lim_{n\to\infty}\cos\frac{1}{n}=1 $$ and therefore $$ \lim_{n\to\infty}n\cos\frac{1}{n}=\infty $$

Note, however, that no rule can be stated when $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n=0$.

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Assume that $\lim_{n}a_{n}=a$ and $\lim_{n}b_{n}=\infty$ for $a\in(0,\infty)$, then given $M>0$, find some positive integer such that $b_{n}>M$ for all $n\geq N$. By choosing sufficiently large $N$, we also have for all $n\geq N$, $|a_{n}-a|<\dfrac{a}{2}$. Then $a_{n}b_{n}>\dfrac{a}{2}\cdot b_{n}>\dfrac{a}{2}\cdot M$, since $M>0$ is arbitrary, this proves $\lim_{n}a_{n}b_{n}=\infty$.

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