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Another sum - convergence

  1. Jan 18, 2005 #1
    Hi,

    I have this sum:

    [tex]
    \sum_{n = 1}^{\infty} \frac{n^2}{ \left( 2 + \frac{1}{n} \right)^{n}}
    [/tex]

    I tried d'Alembert, I tried comparing it with [itex]\frac{1}{2^{n}}[/itex], but without success.

    Could somebody point me to the right direction please?

    Thank you.
     
  2. jcsd
  3. Jan 18, 2005 #2
    [tex] \lim (2+1/n)^n = \lim (2(1+1/2(1/n))^n =\lim 2^n (1+1/n)^{n/2} [/tex]
    can you get from here?
     
  4. Jan 18, 2005 #3
    Thank you vincentchan, but I don't understand the second step - how can I get

    [tex]
    \left(1+\frac{1}{n}\right)^{n/2}
    [/tex]

    from

    [tex]
    \left(1+\frac{1}{2}\left(\frac{1}{n}\right)\right)^{n}
    [/tex]

    I know it must be some simple algebraic adjustment but I can't see that...
     
  5. Jan 18, 2005 #4
    use change of variable, let u=n/2 the limit will become u->infinity/2, but infinity divided by 2 is also infinity... so u->infinity
     
  6. Jan 18, 2005 #5
    OH... i made a mistake in post 2, the big idea is the same... see if you can catch it.....
     
  7. Jan 18, 2005 #6

    Galileo

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    You can also compare it with:

    [tex]\sum_n \frac{n^2}{2^n}[/tex]
     
  8. Jan 18, 2005 #7
    I see it now (with being aware of the mistake you did :) ) I wouldn't see the possibility of the change at the first look, however.
     
  9. Jan 18, 2005 #8
    You're right, this is probably the easiest way. Thank you.
     
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