Comparison test for series convergence (trig function)

[kwal0203]

Homework Statement

Use a comparison test to determine whether this series converges:

$$\sum_{x=1}^{\infty }\sin ^2(\frac{1}{x})$$

Homework Equations

The Attempt at a Solution

At small values of x:

$$\sin x\approx x$$
$$a_{x}=\sin \frac{1}{x}$$
$$b_{x}=\frac{1}{x}$$
$$\lim \frac{a_{x}}{b_{x}}=\frac{\sin \frac{1}{x}}{\frac{1}{x}}=1$$

Since 1/x diverges so does sin(1/x).

Can I use this same method to solve the question above? ( i.e. sin^2(1/x) )

 

[VrhoZna]

Why not just try out that same method? What can you compare sin²(1/x) to?

 

[kwal0203]

Why not just try out that same method? What can you compare sin²(1/x) to?

I guess what I'm not sure about is if this is true:

$$\sin^2 x\approx x^{2}$$

Can I just make that assumption?

 

[VrhoZna]

I guess what I'm not sure about is if this is true:

$$\sin x\approx x^{2}$$

Can I just make that assumption?

No, reckless assumption often leads to errors. Do you recall that special trigonometric limit that involves sin? It involves a limit where x tends to 0 but with a clever use of substitution it may help you.

 

[kwal0203]

No, reckless assumption often leads to errors. Do you recall that special trigonometric limit that involves sin? It involves a limit where x tends to 0 but with a clever use of substitution it may help you.

Hmm, what about this, if:

$$\sin x\approx x$$

then

$$\sin^2 x\approx x^{2}$$

Since I'm just squaring both sides it should still be true right?

Then I can go:

$$a_{x}=\sin ^2\frac{1}{x}$$
$$b_{x}=\frac{1}{x^2}$$

It does work out this way, but what function did you have in mind to compare it to?

 

[VrhoZna]

Hmm, what about this, if:

$$\sin x\approx x$$

then

$$\sin^2 x\approx x^{2}$$

Since I'm just squaring both sides it should still be true right?

Then I can go:

$$a_{x}=\sin ^2\frac{1}{x}$$
$$b_{x}=\frac{1}{x^2}$$

It does work out this way, but what function did you have in mind to compare it to?

You shouldn't use approximations as they don't have a place in mathematical problems of this nature and they don't mean anything without specifying a tolerance for error. sin(pi) is approximately equal to pi like 5 is approximately equal to 2

The limit I was referring to is lim x --> 0 (sinx/x) = 1. Try and figure out its application to this problem.

 

[kwal0203]

You shouldn't use approximations as they don't have a place in mathematical problems of this nature and they don't mean anything without specifying a tolerance for error. sin(pi) is approximately equal to pi like 5 is approximately equal to 2

The limit I was referring to is lim x --> 0 (sinx/x) = 1. Try and figure out its application to this problem.

$$\sin x\leqslant 1$$
$$\frac{\sin x}{x^{2}}\leqslant \frac{1}{x^{2}}$$
$$\frac{\sin^2 x}{x^{2}}\leqslant \sin(\frac{1}{x^{2}})$$

Am I on the right track here?

Sorry, I'm not really sure how sin(x)/x relates to sin^2(1/x) in terms of which is greater than the other. How can I figure that out?