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I'm in trouble with this limit. The numerator diverges positively, but I do not understand how to operate on the denominator.

$$\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})},$$

$$\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})}= \lim_{x\to\infty}\frac {n^2(5 +\frac{\sin n}{n^2})}{3 (n+2)^2 \cos(\frac{n \pi}{5})} \cdots$$

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    $\begingroup$ Show that $\lim\limits_{n\rightarrow\infty}{5n^2+\sin n\over 3(n+2)^2}=5/3$ and observe that ${1\over\cos(n\pi/5)}$ takes on alternating values (1 and $1/\cos(\pi/5)$ e.g.) as $n$ tends to infinity. Can the product of the two converge? $\endgroup$ Nov 20, 2011 at 12:22
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    $\begingroup$ What do you mean by "irregular"? $\endgroup$ Nov 20, 2011 at 12:47
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    $\begingroup$ This is categorized as "real-analysis". Is $n$ integer or real? $\endgroup$ Nov 21, 2011 at 6:15
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    $\begingroup$ @marty: I have removed the (real-analysis) tag. $\endgroup$
    – JavaMan
    Dec 6, 2011 at 16:55
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    $\begingroup$ @FrConnection: I removed the (real-analysis) tag since your posts seem to concern calculus rather than Real Analysis. To gain a better understanding of the topics covered in Real Analysis, you can, for example, look through the questions tagged as (real-analysis): math.stackexchange.com/questions/tagged/real-analysis and you can read more about Real Analysis at en.wikipedia.org/wiki/Real_analysis#Key_concepts, among many other places. $\endgroup$
    – JavaMan
    Dec 6, 2011 at 17:01

1 Answer 1

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Let's make a few comments.

  1. Note that the terms of the sequence are always defined: for $n\geq 0$, $3(n+2)^2$ is greater than $0$; and $\cos(n\pi/5)$ can never be equal to zero (you would need $n\pi/5$ to be an odd multiple of $\pi/2$, and this is impossible).

  2. If $a_n$ and $b_n$ both have limits as $n\to\infty$, then so does $a_nb_n$, and the limit of $a_nb_n$ is the product of the limits of $a_n$ and of $b_n$, $$\lim_{n\to\infty}a_nb_n = \left(\lim_{n\to\infty}a_n\right)\left(\lim_{n\to\infty}b_n\right).$$

  3. If $b_n$ has a limit as $n\to\infty$, and the limit is not zero, then $\frac{1}{b_n}$ has a limit as $n\to\infty$, and the limit is the reciprocal of the limit of $b_n$: $$\lim_{n\to\infty}\frac{1}{b_n} = \frac{1}{\lim\limits_{n\to\infty}b_n},\qquad \text{if }\lim_{n\to\infty}b_n\neq 0.$$

As a consequence of $2$ and $3$, we have:

  • If $\lim\limits_{n\to\infty}a_nb_n$ and $\lim\limits_{n\to\infty}a_n$ exists and is not equal to $0$, then $\lim\limits_{n\to\infty}b_n$ exists:

    Just write $\displaystyle b_n = \left(a_nb_n\right)\frac{1}{a_n}$

  • Equivalently, if $\lim\limits_{n\to\infty}a_n$ exists and is not zero, and $\lim\limits_{n\to\infty}b_n$ does not exist, then $\lim\limits_{n\to\infty}a_nb_n$ does not exist either.

So, consider $$a_n = \frac{5n^2 + \sin n}{3(n+2)^2},\qquad b_n =\frac{1}{\cos(n\pi/5)}.$$ We have, as you did: $$\begin{align*} \lim_{n\to\infty}a_n &= \lim_{n\to\infty}\frac{5n^2 + \sin n}{3(n+2)^2}\\ &= \lim_{n\to\infty}\frac{n^2\left(5 + \frac{\sin n}{n^2}\right)}{3n^2(1 + \frac{2}{n})^2}\\ &=\lim_{n\to\infty}\frac{5 + \frac{\sin n}{n^2}}{3(1+\frac{2}{n})^2}\\ &= \frac{5 + 0}{3(1+0)^2} = \frac{5}{3}\neq 0. \end{align*}$$

What about the sequence $(b_n)$?

If $n=(2k+1)5$ is an odd multiple of $5$, then $$b_n = b_{(2k+1)5}\frac{1}{\cos\frac{n\pi}{5}} = \frac{1}{\cos((2k+1)\pi)} = -1;$$ so the subsequence $b_{(2k+1)5}$ is constant, and converges to $-1$. On the other hand, if $n=10k$ is an even multiple of $5$, then $$b_n = \frac{1}{\cos\frac{n\pi}{5}} = \frac{1}{\cos(2k\pi)} = 1.$$ so the subsequence $b_{10k}$ is constant and converges to $1$.

Since a sequence converges if and only if every subsequence converges and converges to the same thing, but $(b_n)$ has two subsequences that converge to different things, it follows that $(b_n)$ does not converge. (It also does not diverge to $\infty$ or to $-\infty$, since there are subsequences that are constant).

And so, what can we conclude, given our observations above about products of sequences?

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  • $\begingroup$ we can say that the limit of the sequence does not exist! $\endgroup$ Dec 6, 2011 at 19:21

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