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Series problem

  1. Oct 3, 2006 #1
    I found this problem in Rudin, and I just can't figure it out.

    It goes like this:
    Prove that the convergence of [tex]\sum a_n[/tex] [tex] a_n \geq 0[/tex] implies the convergence of [tex] \sum \frac{\sqrt{a_n}}{n} [/tex]

    I tried the comparison test, but that doesn't help because I don't know what the limit [tex]\lim_{n \rightarrow \infty} \frac{1}{n\sqrt{a_n}}[/tex] is equal to.

    Then I tried the partial summation formula, [tex] \frac{1}{n} \rightarrow \infty [/tex] and is monotonic, but [tex]\sqrt{a_n} > a_n[/tex] for all but finite many n. [tex]\sqrt{a_n}[/tex] is rising, so if it the partial sums were bounded from above the series would converge, but that isn't true for [tex] a_n = \frac{1}{n^2} [/tex], so I can't use this way.

    The last thing that comes to my mind is to use the Cauchy criterion, but I can't find any good use of it here. a_n will be smaller than any epsilon for infinitely many n, but that doesn't really help.

    Have I missed something out, or done something in a wrong way? Thanks for any help.
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2


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    When you tried partial summation, what kind of bounds did you use for [tex]\sum_{n\leq x}\sqrt{a_n}[/tex]?
  4. Oct 4, 2006 #3
    The problem is there are no bounds for it - like in the example I mentioned - [tex]\sum a_n = \sum \frac{1}{n^2} [/tex] converges, but [tex] \sum \sqrt{a_n} = \sum \frac{1}{n} [/tex] doesn't, hence it is also not bounded from above.
    Last edited: Oct 4, 2006
  5. Oct 4, 2006 #4


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    I don't mean bounded by a constant for all partial sums, rather an upper bound that depends on x.

    With [tex]a_n=1/n^2[/tex], you can get the bound [itex]\sum_{n\leq x}\sqrt{a_n}\leq \log(x)+1[/itex]. Using this bound, you could go throught the partial summation and show [itex]\sum_{n\leq x}\sqrt{a_n}n^{-1}[/itex] is bounded by a constant.

    So, what kind of bounds can you get for [itex]\sum_{n\leq x}\sqrt{a_n}[/itex]? It would be good to ask what kind of bound you'd need to deduce convergence from partial summation as well.
    Last edited: Oct 4, 2006
  6. Oct 4, 2006 #5


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    How about the "root test"? Since [itex]\Sigma a_n[/itex] converges, you know that [itex]^n\sqrt{a_n}[/itex] converges to a number less than or equal to one. What does that tell you about the limit of
    [tex]{^n\sqrt{\frac\sqrt{a_n}}{n}}= \frac{\sqrt{^n\sqrt{a_n}}}{^n\sqrt{n}}[/tex]?
  7. Oct 4, 2006 #6
    the root test!!!! I don't know why, but I just ignored it. Thanks a lot
  8. Oct 4, 2006 #7


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    The root test will be inconclusive here if you started with a sequence that has [tex]\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=1[/tex] to begin with. Or if this limit simply didn't exist (though you can use the lim sup version, again inconclusive if you get a "1").
  9. May 4, 2008 #8
    If [tex]
    a_n \le {1 \over {n^2 }}
    [/tex], then [tex]
    \sqrt {a_n } \le {1 \over n} \Rightarrow {{\sqrt {a_n } } \over n} \le {1 \over {n^2 }}

    if [tex]
    a_n \ge {1 \over {n^2 }}

    [/tex], then [tex]
    \sqrt {a_n } \ge {1 \over n}

    [/tex]; multiplying both sides of this inequality by [tex]
    \sqrt {a_n }

    [/tex] we obtain [tex]
    {{\sqrt {a_n } } \over n} \le a_n


    Thus [tex]
    {{\sqrt {a_n } } \over n} \le \max \left\{ {a_n ,{1 \over {n^2 }}} \right\} = {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) + {1 \over 2}\left| {a_n - {1 \over {n^2 }}} \right| \le {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) + {1 \over 2}\left( {a_n + {1 \over {n^2 }}} \right) = a_n + {1 \over {n^2 }}


    Then the convergence of [tex]
    \sum\limits_n {a_n }

    [/tex] and [tex]
    \sum\limits_n {{1 \over {n^2 }}}

    [/tex] implies the convergence of

    \sum\limits_n {{{\sqrt {a_n } } \over n}}

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