Recent content by jokerzz

the answer's 0?

no there's no x^2. When you put x=0 at the end after substituting the mclaurin series of e^x and sinx, everything becomes 0, except -1/3! + 1/2 which is equal to 1/3

Lim x-> 0 (e^xsin(x)-x-x^2)/(x^3 +xln(1-x)) I first put the Mclaurin series of e^x and sinx and then took x^3 common and then put the limit. My answer comes out to be 1/3. Can anyone confirm whether I've done it right?

Finally I get it! Thanks a lot dude!

Yeah all the terms accept -ln2 are cancelling out. But isn't this method a little crude? Isn't there any other way to prove it?

Im sorry but my concepts in this topic arnt very clear. Can you please elaborate a little?

Ok so now the equation is: ln(k-1)+ln(k+1) - 2ln(k) so now what do i need to do? Apply the limit k-> infinity? But that wouldn't prove anything

sigma k=2 to infinite is also there

I have to show that ln(1 - 1/k^2)= - ln(2) I took LCM and separated the equation using ln(A/B) = lnA - lnB How should i proceed now?

So ur saying for question 2 its trial and error? Isnt there any method like a(n+1)-an<0 or sumthing?

your answer makes no sense to me! I don't think you have to be soooo damn cryptic

I have 2 questions. How do you use differetiation to prove whether sequence is monotonic? For example: 1/n+ln(n) My 2nd question is, how do you prove whether sequence is EVENTUALLY monotonic?

Thanks dude for your help. I understand now

The question I originally asked was just part of the whole question. Read my post above

It is?? How come? I took it as "(2/3)^n"