公主的记事本

3, 1, 1/3, ...

Show that the G.P is convergent or divergent.

Find the sum to infinity.

* I wrote the question based on my memory... So it maybe a bit different from the actual.

The first part I don't think need to explain... Because what I want to talk about is part 2.

Before I move on, let me explain how she solved the question. At first, she tell us the that the sequence is convergent. Suddenly, a student asked, why it is convergent. Funny, she stunned and turn to the white board and tried to solve the prove it secretly... She writing all the lim (n approach infinity)... She refer to notes... Then come out with result that the sequence is divergent... Then is my turn to stunned!

Well, based on our previous lecturer, he told that when the sequence is getting smaller and smaller, at the end it will stop at 0 or any other number. Therefore, the sequence is convergent. Divergent is when the sequence getting larger and larger and it will not stop on a specific number but keep on growing.

Ok, back to the question, the sequence is, obviously, decreasing. So, its definitely that this sequence is convergent. I told her the same thing (I even bring our previous lecturer into my answer!). What she reply? Gosh... she said it is explanation and looked like she is not accepted it.

Then I said, if the sequence is divergent, we can't even find the sum to infinity, but we had found it! It proved that the sequence is convergent! Then she answer me that is the matter of the order of question...

Fine! Then we stick to her method. The nth term is =3(1/3)^n-1

Lim_n->∞ 3(1/3)^n-1 = 3 . Lim_n->∞ (1/3)^n-1

Which Lim_n->∞ (1/3)^n-1 = 0

Some may asked, why Lim_n->∞ (1/3)^n-1 = 0

Well, 1/3 and (1/3)² ,which is smaller? Obviously, is (1/3)²

Then when the n(the power) keep on growing, the number will get more closer to 0 and finally stop at 0. Therefore, Lim_n->∞ (1/3)^n-1 = 0