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mathaddict4471
challenge calculus question without a calculator use the Maclaurin series to compute e^1 medal and fan rewarded
\[\lim_{n \rightarrow \infty} n \sin (2pien!)\]
sry compute equation above using e^1
Is the expression n.sin(2.pi.e.n!) ?
...which can't be right. What exactly are we taking the limit of?
the first thing you said
we're taking the limit of n sin i think
i'll use wolfram
http://www.math.utah.edu/ugrad/calc_challenge/calc-challenge-2006.pdf it's number 6
I see. Ok, let me think.
@mathaddict: the equation must be e^x , right? to apply Maclaurin series, how can you apply to e^1
The problem actually says: Using the Maclaurin series for \( e^1 \), calculate \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]
Ok. This is what I've got so far. When n = 2, n!.e = 2!(1 + 1 + 1/2 + 1/3! + ...) = 2 + 2 + 1 + 2/3! + ... Now sin(2.pi.e.n!) = sin(2.pi ( 2 + 2 + 1 + 2/3! + ... ) ) = sin(2.pi ( 5 + stuff) ) Recall that sin(a + b) = sin a . cos b + cos a . sin b hence sin(2.pi.e.n!) = sin(2.pi.5).cos(stuff) + cos(2.pi.5).sin(stuff) = 0 .cos(stuff) + 1 .sin(stuff) = sin(stuff) This argument generalizes to \[ \sin(2\pi e n!) = \sin \left[ 2\pi \left( \sum_{i=0}^n n!/i! + \sum_{i=n+1}^{\infty} n!/i! \right) \right] = \sin \left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Thus the limit is equal to this limit \[ \lim_{n \rightarrow \infty} n . \sin\left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Now we want to figure out how to bound the quantity in square brackets and show it goes to zero sufficiently fast to cancel out the n outside altogether or end up with some finite number
@JamesJ : are you Math Professor?
In fact that sum is 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... < 1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ... and this infinite geometric series is equal to 1/n Hence the sum is bounded above by the limit as n --> infinity of n.sin(2pi/n) = sin(2pi/n)/(1/n) --> 2pi So I imagine the limit is in fact 2pi. Now a little more work is required to show that it is exactly 2pi
Oh, and that's easy because 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... > 1/(n+1) and hence the limit is bounded below by the limit of n.sin(2pi/(n+1) which has the same limit. **** Nice question! Yes, I was once.
I love your neatness. It means you used to be a Math Professor, right? (sorry for Capitalize the P in professor, because to me, those people deserve that C)
This is very counter-intuitive result. Adding this to the the archives.
Can I make question when i have some via message? I don't want to post and wait wait wait..... until someone have free time .
Better to post the question and then mail me with the link.
how to post the link. once , other asked me to do that but I didn't know how to send the link
Just copy it from your browser and paste it into a message
do I have to be your fan to send the message? I try reply yours and send message to your profile. both are fail. I cannot submit the mail
Hm, possibly. I'm not sure.
thank you soo much for your greatly intuitive answer James, you are the master. best of luck to both of you JamesJ and Hoa