# Calculus

Arachnarchist
Posers and Puzzles 15 Oct '07 17:10
1. 15 Oct '07 17:10
Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
2. PBE6
Bananarama
15 Oct '07 17:29
Originally posted by Arachnarchist
Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
We can rearrange this sequence and make a simple substitution to make it more recognizable:

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.
Baby Gauss
15 Oct '07 17:322 edits
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.

4. TheMaster37
Kupikupopo!
15 Oct '07 17:46
Originally posted by Arachnarchist
Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
As for n -> 0

sin(1/n) is bounded, so n*sin(1/n) -> 0
Baby Gauss
15 Oct '07 17:56
Originally posted by TheMaster37
As for n -> 0

sin(1/n) is bounded, so n*sin(1/n) -> 0
It's n*sin(1/n) not 1/n*sin(1/n).
The first time I also read this wrong though not like you
6. 15 Oct '07 18:29
Originally posted by TheMaster37
As for n -> 0

sin(1/n) is bounded, so n*sin(1/n) -> 0
I think he meant n -> infinity, not n -> 0. I'll go with what PBE6 said.
7. 15 Oct '07 18:30
Sequence v Series ðŸ™‚
8. TheMaster37
Kupikupopo!
15 Oct '07 20:10
It's n*sin(1/n) not 1/n*sin(1/n).
The first time I also read this wrong though not like you
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
Baby Gauss
15 Oct '07 21:16
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.
10. genius
Wayward Soul
19 Oct '07 07:58
Originally posted by Arachnarchist
Determine whether the following sequence converges or diverges. If it converges find the limit. an=n(sin(1/n))
why "calculus"?
11. PBE6
Bananarama
19 Oct '07 14:11
Originally posted by genius
why "calculus"?
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.
12. genius
Wayward Soul
19 Oct '07 15:12
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*
13. sonhouse
Fast and Curious
27 Oct '07 13:43
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™‚
14. AThousandYoung
All My Soldiers...
27 Oct '07 18:04
Originally posted by sonhouse
What does PBE6 mean? You want 1,000,000 Pall Bearers?ðŸ™‚
Pirate Banana Evildoer #6.

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