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watchman can you check this and tell me what you think (i have to attach one sec)
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oops i said watchman i meant watchmath lol
hey watchmath i assumed e was a good approximate for \[\sum_{i=0}^{2011}\frac{i}{i!}\]
i have to leave but i might be back feel free to say anything i will read it :)
oh wait i wait
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The integral is divergent
ok i will look at it some more later and hopefully i can come up with that conclusion
did you look at my attachment?
@myininaya: I think the integral should be from 1 to infinity.
oops you are right lol
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The answer is 2011!/(2010)^(2012)
Use this: \(\int_0^{\infty}e^{-t} t^n \, dt =n!\)
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