Recent content by Watson91

Yeah blocking is a problem for me when it comes to calculus. I seem to get hung up on things that shouldn't be a problem for myself. I'm not sure how to overcome the issue though other than just asking for help.

Yep the question is of the sequence. Thanks for helping me out on that problem, even though I drug it on. It really wasn't that difficult after I look back over it, but it always seems that way for me. I get a mental block and once I get past it I question why I couldn't solve it originally...

What the question asks is to determine convergence or divergence. Maybe I am confusing the two. I think what would answer my question is; if you are able to find the limit then your sequence converges, I think this is correct for a sequence. To my knowledge the limit of 1/2n = 0, so if...

Well I know that the lim of 1/n = 0 and diverges. By comparison I would say that 1/2n > 1/n which would make me believe it diverges. Would this be correct?

Great, now back to my original question. How do I decide if the limit converges or diverges from here? This is the part I'm really stuck/confused about.

I understand your concern now. I used what I thought was the chain rule by taking the derivative of the outside ( 1/n ) and then multiplied by the derivative of in the inside ( 1/2( sqrt(n) ). Really I did the problem incorrectly I can see because what I tried to say was n X sqrt(n) = n...

Derivative = 1/n X 1/2\sqrt{n} = \frac{1}{2n}

If I'm not mistaken the derivative of the ln \sqrt{n} = \frac{1}{2n} Please correct me if I'm wrong, I'm quite unsure of my work at this point to be honest.

Homework Statement I have been asked to determine convergence or divergence of a sequence given the nth term. If the sequence converges, find its limit. an = ln\sqrt{n} / n Homework Equations The Attempt at a Solution I had thought that the sequence would follow L'Hopital's...