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Bounding Argument for Comparison Test

  1. Jun 3, 2015 #1
    I have the sum,
    $$\sum_{n=1}^{\infty} \frac{1}{n^{3}}\sin(n \pi x) \text{, where }0 \leq x \leq 1$$
    I have to show that the series converges, so I'm going to use the Comparison Test.
    $$ \text{If }0 \leq a_n \leq b_n \text{ then}$$$$\text{If }\sum b_n \text{ converges then }\sum a_n \text{ must converges.}$$
    Applying the Comparison Test, I know I must show that,
    $$0 \leq \frac{1}{n^{3}}\sin(n \pi x) \leq \frac{1}{n^3} \text{, where }0 \leq x \leq 1$$
    (This is hard part), Since we know that x is between 0 and 1, I think we can bound it.
    $$0 \leq \frac{1}{n^{3}}\sin(n \pi (0)) \leq \frac{1}{n^{3}}\sin(n \pi (1)) \leq \frac{1}{n^3}$$
    Since 0 times anything is 0, and 1 times anything is 1, we get
    $$ 0 \leq \frac{1}{n^{3}}\sin(0) \leq \frac{1}{n^{3}}\sin(n \pi ) \leq \frac{1}{n^3}$$
    Since sine of 0 is 0, we get
    $$ 0 \leq 0 \leq \frac{1}{n^{3}}\sin(n \pi ) \leq \frac{1}{n^3}$$
    And since sine of n times pi alternates between sine of 1 pi and 2pi, sine equals 0.
    $$ 0 \leq 0 \leq 0 \leq \frac{1}{n^3}$$
    And since this holds true, and we know that 1 divided by n cubed converges due to the p-series, the sum must therefore converge.

    I feel like this isn't correct, because if x = 1/2 and n = 3, we get sine of 3pi/2, which equals -1 and then this falls apart. If possible can someone nudge me in the right direction? Much thanks!
  2. jcsd
  3. Jun 4, 2015 #2


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    Indeed the term you are summing are not always positive, so your proof is incorrect (your argument about ## \sin(n\pi x) ## is not justified, sine is not an increasing function)

    Instead, try to show that your series is absolutely convergent.
  4. Jun 4, 2015 #3


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    Basis of proof:
    [itex]0\le \frac{|sinx|}{n^3} \le \frac{1}{n^3}[/itex]
  5. Jun 4, 2015 #4
    I didn't even think about about absolute convergence, since ratio and root test failed. I think I can take it from here, I just gotta show that it is absolutely convergent using the comparison test and then that will be it. Much Thanks!
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