I have the sum,(adsbygoogle = window.adsbygoogle || []).push({});

$$\sum_{n=1}^{\infty} \frac{1}{n^{3}}\sin(n \pi x) \text{, where }0 \leq x \leq 1$$

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!

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# Bounding Argument for Comparison Test

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