This question has been asked in:
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."