Originally posted by Arachnarchist Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))

Well the limit of the general term when n tends to infinity is
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 adam warlock 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 TheMaster37 One thing you learn in time is to read carefully. It makes life so much easier at times ðŸ˜‰

Tohugh I occasionally muck up as well :p

LOL
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 ðŸ˜³. Automod was on full power.

Originally posted by PBE6 Because you have to take the limit as "n" tends toward infinity to see if the sequence converges or not. That's some straight-up calculus, yo.

*Bows his head in shame as he re-learns what he thought the Calculus was*

Originally posted by PBE6 Because you have to take the limit as "n" tends toward infinity to see if the sequence converges or not. That's some straight-up calculus, yo.

What does PBE6 mean? You want 1,000,000 Pall Bearers?ðŸ™‚