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Is this the right approach to solve it?

  1. Dec 1, 2004 #1
    Hi all,

    my task is to solve the convergence of the sum in dependance to the parameter a real.

    [tex]
    \sum_{n=1}^{\infty} \frac{a^{n}}{1+a^{n}}
    [/tex]

    I did it this way:

    First I found out that if the sum converges, a will have to be in interval (-1, 1). So how to prove that for these values of parameter the sum converges?

    Let's try d'Alembert's criterion, which tells us this:

    [tex]
    \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} < 1 \Rightarrow \sum a_{n}{} is convergent
    [/tex]

    So:

    [tex]
    \lim_{n \rightarrow \infty} \frac{a^{n+1}}{1+a^{n+1}} : \frac{a^{n}}{1+a^{n}} =
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} \frac {a^{n+1}.(1+a^{n})}{a^{n}.(1+a^{n+1})} =
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} \frac {a.(1+a^{n})}{1+a.a^{n}} =
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} \frac {a}{a^{n+1} + 1} + \lim_{n \rightarrow \infty} \frac{a^{n+1}}{a^{n+1} + 1}
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} \frac {a}{a^{n+1} + 1} = \lim_{n \rightarrow \infty} \frac { \frac{a}{a}}{a^{n} + \frac{1}{a}} = \frac{1}{0 + \frac{1}{a}} = 0
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} \frac {a^{n+1}}{a^{n+1} + 1} = \lim_{n \rightarrow \infty} \frac {1}{1 + \frac{1}{a^{n+1}}} = 0
    [/tex]

    Thus

    [tex]
    \lim_{n \rightarrow \infty} \frac {a^{n+1}}{a^{n+1} + 1} : \frac{a^{n}}{1+a^{n}} = 0 + 0 = 0
    [/tex]

    This way I proved the convergence for a in (-1, 1)

    Is it ok?

    Thank you for your comments.
     
  2. jcsd
  3. Dec 1, 2004 #2

    arildno

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    That limit is incorrect, the expression goes to "a", not to 0.
    However, since a<1, convergence is achieved.
     
  4. Dec 1, 2004 #3
    You're right, I didn't take into account that a is constant.
     
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