Bounding Argument for Comparison Test

Main Question or Discussion Point

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!

wabbit
Gold Member
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)

mathman
$0\le \frac{|sinx|}{n^3} \le \frac{1}{n^3}$
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)