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Math Help - finding values for convergence

  1. #1
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    finding values for convergence

    I'd like to find for which values of x does
    (1) sum n=1 to infinity (x^n)/n converge?
    (2) sum n=1 to infinity (x^n)/(n^2) converge?
    (3) sum n=1 to infinity (n^x)/(n^2 +1) converge?

    Do i all do the ratio test and set the limit of it less than 1??? or is there different rules to consider for the each cases???

    for instance for the (1) I took the ratio test and then I set the limit of it less than 1
    lim n->oo (x^n+1)/(n+1) <1 which then becomes (x^n+1) has to be less than n+1
    then x must be less than (n+1)^-(n+1) is this the right way of solving the problem??? help !!
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  2. #2
    Super Member girdav's Avatar
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    Be careful when you apply the ration test. For example, in the first question, put a_n(x) =\frac{x^n}n. We have \left|\frac{a_{n+1}(x)}{a_n(x)}\right|=\frac{|x|n}  {n+1}. Now you have to take the limit n\to\infty (look at the proof if it's not clear).
    The ratio test works for the second question, for the third find the x such that the limit when n\to\infty is 0.
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  3. #3
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    Quote Originally Posted by girdav View Post
    for the third find the x such that the limit when n\to\infty is 0.
    Be careful indeed.
    Notice that if x=1 then \frac{n^1}{n^2+1}\to 0.
    But \sum\limits_{n = 1}^\infty  {\frac{n}{{n^2  + 1}}}  \geqslant \sum\limits_{n = 1}^\infty  {\frac{1}{{2n}}}
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  4. #4
    Super Member girdav's Avatar
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    Quote Originally Posted by Plato View Post
    Be careful indeed.
    Notice that if x=1 then \frac{n^1}{n^2+1}\to 0.
    But \sum\limits_{n = 1}^\infty  {\frac{n}{{n^2  + 1}}}  \geqslant \sum\limits_{n = 1}^\infty  {\frac{1}{{2n}}}
    I know. I gave this "hint" because it's not easy to see the convergence without any computation (but the problem is that in this case the might be a lot of x such that a_n(x)\rightarrow 0 and the series is not convergent). I realize it's not the shortest way to solve it. If x\leq 0 the series is convergent, and if x>0 the series has the same behavior as \sum_{n=1}^{+\infty}\frac 1{n^{2-x}}.
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