Originally posted by ArachnarchistWe can rearrange this sequence and make a simple substitution to make it more recognizable:
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.
2 edits
Originally posted by ArachnarchistWell the limit of the general term when n tends to infinity is
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.
Originally posted by TheMaster37LOL
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.
