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- Apr 15th 2009, 04:05 AMm3th0dmanLimit when n-> infinity

- Apr 15th 2009, 08:10 AMOpalg
- Apr 15th 2009, 12:22 PMm3th0dman
Thank you very much. This is the correct answer (written at the end of the Book). Unfortunately I am not familiar with Big O notation, so I will have to chew on it a little.

I would be grateful if you could explain it without "Big O" notation, or if easier you could explain the "Big O" notation.

I understand that is a function that is not important in finding the answer, written to simplfy things.

In this is case it depends on 1/n^2, so I supouse it goes to 0 when n-> infinity. - Apr 16th 2009, 02:15 AMOpalg
The big O notation is just a succinct way of estimating orders of magnitude. For this problem, you can make the estimate explicit by using Taylor's theorem with remainder for the function ln(1+x). This tells you that, for 0<x<1, the difference between ln(1+x) and x is less than . Put to see that the difference between and is less than . Adding n+1 such terms together, you find that the difference between and is less than . This difference goes to 0 as n goes to infinity.

- Apr 16th 2009, 04:38 AMm3th0dman
I suppose that Taylor's formula is applied this way:

And we apply it for a=0:

and

for x between [0,1].

I haven't studied Taylor's Theorem (or formula) at school, so I don't know if I had applied correctly.

Assuming what have I written is correct, I understood everything. Again I thank you very much.