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Need series convergence help

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Does the series
    [tex]
    \Big( \sum_{n=1}^\infty\frac{1}{(3^n)*(sqrtn)} \Big)
    [/tex]
    Converge or Diverge? By what test?


    2. Relevant equations
    1/n^p
    If p<1 or p=1, the series diverges.
    If p>1, the series converges.

    If bn > an and bn converges, then an also converges.

    3. The attempt at a solution
    I use 1/(sqrtn) since it is bigger than 1/((3^n)(sqrtn)).
    Since sqrtn is n^1/2 I use the p test.
    Since 1/2<1, the series 1/sqrtn diverges and so does the original.
    This is wrong

    The answer uses 1/3^n as the comparison and it just says that it converges with no explanation.
    Also, the ratio test was used and it converged.

    Why doesn't my reasoning work?
     
  2. jcsd
  3. May 19, 2013 #2

    haruspex

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    You showed that a series with larger terms, and therefore less inclined to converge, doesn't converge. What can you deduce from that?
     
  4. May 19, 2013 #3

    CAF123

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    Gold Member

    You are misapplying the Comparison Test. It is correct to state that $$\frac{1}{3^n \sqrt{n}} < \frac{1}{\sqrt{n}},$$ however while the the sum of the RHS diverges (by p test), you cannot conclude anything about the convergence/divergence of LHS. As a counterexample to your line of reasoning , see for example the following:
    $$n^2 > n\,\,\text{for}\,\,n > 1 \Rightarrow \frac{1}{n^2} < \frac{1}{n}.$$ The infinite sum of RHS diverges (by p test) while the LHS converges (by p test).
     
  5. May 19, 2013 #4
    Hi

    $$ \sum_{n=1}^\infty\frac{1}{3^n\cdot\sqrt{n}} < \sum_{n=1}^\infty\frac{1}{3^n}=\frac{1}{2} $$
     
    Last edited: May 19, 2013
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