Alex Miller
Alex
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| Posted on Thursday, 30
January, 2003 - 11:28 pm: |
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Lim
N*Sin[2*pi*e*N!] as N tends to infinity. can anyone
please help. I must be missing something easy.
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Alex Miller
Alex
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| Posted on Friday, 31
January, 2003 - 12:10 am: |
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It goes to
2*Pi but how?
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David
Loeffler
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| Posted on Friday, 31
January, 2003 - 09:13 am: |
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Try substituting in the series
expansion for e, that is
You should find that you get a lot of integer terms,
which can be thrown away as sin x=sin(x+2pn), a term
1/(n+1) and lots of terms that are too small to matter.
What you get is thus asymptotically
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N sin |
æ
ç
è
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2p
N+1
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+O( |
1
n2
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) |
ö
÷
ø
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~ |
2pN
N+1
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~ 2p |
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David
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Alex Miller
Alex
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| Posted on Friday, 31
January, 2003 - 09:10 pm: |
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thank you
very much.
-Alex Miller
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Alex Miller
Alex
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| Posted on Friday, 31
January, 2003 - 09:14 pm: |
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By the way
I really like your site; you have done some beautiful
problems.
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