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Does the sequence converge or diverge? $an+ n[\sin(\pi/n)]$

That post/title is messy to understand, so please refer to the problem as stated in my title.

I don't understand the answers and am unable to comment, so I had to re-ask the question here. In the first answer, it is stated:

"It diverges, but for a different reason. sinx < x for x∈(0,∞)" Why is this true?

It then states, "sinx≥0 for x∈(0,π/2]" I understand this is true, but do not see how this is applied once the inequality is set up for the squeeze theorem.

My own personal work is as follows:

-1 <= sin (pi/n) <= 1
-n <= n sin(pi/n) <= n
using the range of sin(x), then multiplying by n.

The limits of both sides as n--> inf are not the same, so I'm unable to proceed from here.


NOTE: The textbook answer is given as "pi."

  • $\begingroup$ Forgot to add that the answer is "pi" according to the textbook. I will edit this in, as well. $\endgroup$
    – Spag Guy
    Apr 9, 2014 at 6:58

4 Answers 4


Rewrite as $$\lim_{n\rightarrow \infty}\pi \frac{\sin(\pi/n)}{\pi/n}$$

We know that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ so that is enough to determine that the limit exists.


You can geometrically show the result. Its equal to $\pi$. Think about a regular polygon inscribed in a circle. The perimeter of the polygon is $n*s$, where $n$ is the number of sides the polygon has, and $s$ is the side length of the polygon. $s$ is related to $r$, the circles radius, by $\frac{s}{2} = r*\sin(\frac{2\pi}{2n})$. This comes from making a triangle using two adjacent vertices and the center. Substitute in, and notice that as $n$ goes to $\infty$, the polygon becomes the circle. Hence the result

$$2\pi r \approx s*n = 2r * n*\sin\left(\frac{\pi}{n}\right) $$ $$ \rightarrow \pi = \lim_{n\to\infty} n*\sin\left(\frac{\pi}{n}\right) $$

Hope this helps.



$n \mapsto \infty$ , $\frac{\pi}{n} \mapsto 0^+$

and so does its $\sin$.

Hence, it is indeterminate $\infty * 0^+ $. How about rearranging and L Hospital rule?


We will use the following 2 facts:

  1. For $x<10^\circ, \sin x\approx x$
  2. As $n\to\infty$, $\frac{\pi}{n}\to0$

Fact 2 implies that, for large $n$, $\sin \frac{\pi}{n} \approx \frac{\pi}{n}$. Then $$\lim_{n\to\infty} (n\cdot\sin(\frac{\pi}{n})) = \lim_{n\to\infty} (n\cdot\frac{\pi}{n}) = \lim_{n\to\infty} \pi = \pi$$


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