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Series Limit

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Series k=1 to k=Infinity of:


    a) e
    b) 2e
    c) (1+e)(e-1)
    d) e^2
    e) Infinity

    2. Relevant equations

    e^x = Series from n=1 to n=Infinity of:
    x^n / n!

    3. The attempt at a solution

    I was guessing this would end up being C but the answer is infact B. As far as I can tell 2e is equal to Series 2^k / k!, I'm unsure how this is equal.
  2. jcsd
  3. Oct 13, 2008 #2


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    Homework Helper

    You don't need to get to x^n/n!, but to 1/n!, since this series gives e. A start:

    \frac{{k^2 }}{{k!}} = \frac{{kk}}{{k\left( {k - 1} \right)!}} = \frac{k}{{\left( {k - 1} \right)!}}
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