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Help in infinite series problem

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that:
    [n+1 / n^2 + (n+1)^2 / n^3 + ... + (n+1)^n / n ^ (n+1) -> e-1

    2. Relevant equations

    I have been trying it for couple of days. Tried to work the terms, natural log it all, use the byniomial theory but i can´t get to the right answer.

    3. The attempt at a solution

    Working the terms:

    n+1/n^2 = n( 1+1/n) / n^2 = (1+1/n)/n
    (n+1)^2/n^3 = n^2 (1+1/n) ^2 / n^3 = (1+1/n)^2/n

    Ok, so at n:

    (n+1)^n / n^(n+1) = n^n (1+1/n) ^ n / n ^(n+1) = (1+1/n) ^n / n

    Adding the terms up:

    [(1+1/n) + (1+1/n)^2 +... (1+1/n)^n] / n

    Adding the sequence: ( a + ar^2 + ar^3 + ar^n = a / 1 - r)

    (1/n) / [1 - (1+1/n)] = (1/n) / -1/n = -1

    Second try using logarithm got to:

    1 - [ ln (1+1/n)^n / ln n ] and couldn,t progress further

    Thank you very much in advance for any help provided.
     
  2. jcsd
  3. May 28, 2014 #2

    jbunniii

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    Science Advisor
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    Gold Member

    Looks OK up to here.

    No, you're only adding finitely many terms. If we put ##a = 1 + 1/n##, you have
    $$\frac{1}{n}(a + a^2 + a^3 + \cdots + a^n) = \frac{a}{n}(1 + a + a^2 + \cdots + a^{n-1}) = \frac{a}{n}\left(\frac{1 - a^n}{1-a}\right)$$
    Now plug in ##a = 1 + 1/n## and simplify.
     
  4. May 28, 2014 #3
    Thank you very very much.

    I understand it now and finally solved it.

    Thank you again!!
     
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