# Bounding Argument for Comparison Test

1. Jun 3, 2015

### Sai-

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. Jun 4, 2015

### wabbit

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)

3. Jun 4, 2015

### mathman

Basis of proof:
$0\le \frac{|sinx|}{n^3} \le \frac{1}{n^3}$

4. Jun 4, 2015

### Sai-

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!