- 15 Oct '07 17:29

We can rearrange this sequence and make a simple substitution to make it more recognizable:*Originally posted by Arachnarchist***Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))**

a(n) = n*sin(1/n) = sin(1/n)/(1/n)

Let x = 1/n. Then we have:

a(x) = sin(x)/x

As n->inf, x->0, so we must find lim(x->0) a(x). This is easily done using L'Hopital's rule:

lim(x->0) sin(x)/x = lim(x->0) cos(x)/1 = 1

So this sequence does converge to a value of 1. - 15 Oct '07 17:32 / 2 edits

Well the limit of the general term when n tends to infinity is*Originally posted by Arachnarchist***Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))**

lim an = lim n*sin(1/n) = lim n/n = 1.

Since the general term doesn't tend to zero we can say that the series diverges.

Edit: Somehow I read your post as a question if the series of general term n*sin(1/n) converges. Sorry about that. Having read PB's answer I realised that I read it wrong. - 15 Oct '07 21:16

LOL*Originally posted by TheMaster37***One thing you learn in time is to read carefully. It makes life so much easier at times**

Tohugh I occasionally muck up as well :p

The thing is that I'm only used to calculating limits as n goes to infinity is this types of exercises. I think now that I didn't even read your post . Automod was on full power.