Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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

share|cite|improve this question
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? – David Mitra Nov 20 '11 at 12:22
What do you mean by "irregular"? – David Mitra Nov 20 '11 at 12:47
This is categorized as "real-analysis". Is $n$ integer or real? – marty cohen Nov 21 '11 at 6:15
@marty: I have removed the (real-analysis) tag. – JavaMan Dec 6 '11 at 16:55
@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): and you can read more about Real Analysis at, among many other places. – JavaMan Dec 6 '11 at 17:01
up vote 3 down vote accepted

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?

share|cite|improve this answer
we can say that the limit of the sequence does not exist! – FrConnection Dec 6 '11 at 19:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.