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Math Help - Finding the limit as n goes to infinty

  1. #1
    ein
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    Finding the limit as n goes to infinty

    I've been trying to puzzle out how to find this limit but I am well and truly stumped. I would really appreciate the help...

    Find the limit as n goes to infinity of (e^(1/n)+e^(2/n)+e^(3/n)+........+e^(n/n))/n


    Thanks!
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  2. #2
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    Quote Originally Posted by ein View Post
    I've been trying to puzzle out how to find this limit but I am well and truly stumped. I would really appreciate the help...

    Find the limit as n goes to infinity of (e^(1/n)+e^(2/n)+e^(3/n)+........+e^(n/n))/n


    Thanks!

    HI ,

    If you have learnt a topic called Riemann Sum , I advice to apply it to obtain the limit .

    If you not or if you feel bored about this method , i suggest solving it in this way :

     \lim_{n\to\infty} \frac{1}{n} (e^{1/n} +  e^{2/n} +  e^{3/n} + ... + e^{n/n} )

     = \lim_{n\to\infty} \frac{1}{n} (t +  t^2 +  t^3 + ... + t^n ) where  t = e^{1/n}

     = \lim_{n\to\infty} \frac{1}{n} \frac{t( t^n - 1)}{t-1}

    Since  \lim_{n\to\infty}  e^{1/n} ( or ~ t)  = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^{n/n} =\lim_{n\to\infty} (1 + \frac{1}{n} )

    The limit is

    \lim_{n\to\infty} \frac{1}{n} (1+\frac{1}{n}) \frac{e-1}{1+\frac{1}{n} - 1 }

     = \lim_{n\to\infty} \frac{1}{n} n(1+ \frac{1}{n})(e-1)

     = \lim_{n\to\infty} (1+\frac{1}{n})(e-1) = e-1
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  3. #3
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    Another alternative is to realise that the numerator is a geometric series.

    Writing it in closed form, i.e. \frac{a(1 - r^n)}{1 - r} will simplify matters greatly.
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