# Comparison test for series convergence (trig function)

• kwal0203
In summary: 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?I don't think so. In what ways are they different?I don't think so. In what ways are they different?
kwal0203

## Homework Statement

Use a comparison test to determine whether this series converges:

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

## 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) )

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

VrhoZna said:
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?

kwal0203 said:
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.

VrhoZna said:
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.

$$\sin x\approx x$$

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?

kwal0203 said:

$$\sin x\approx x$$

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.

VrhoZna said:
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?

## What is the Comparison Test for series convergence?

The Comparison Test is a method used to determine the convergence or divergence of a series by comparing it to a known series with known convergence behavior. It can be used for series with positive terms.

## How does the Comparison Test work?

The Comparison Test states that if the terms of a series are less than or equal to the terms of a convergent series, then the original series is also convergent. Conversely, if the terms of a series are greater than or equal to the terms of a divergent series, then the original series is also divergent.

## What types of series can the Comparison Test be used for?

The Comparison Test can be used for series with positive terms, including series with trigonometric functions such as sin(x) and cos(x).

## Are there any limitations to the Comparison Test?

Yes, the Comparison Test can only be used for series with positive terms. It also cannot determine the exact value of the series, only whether it converges or diverges.

## Can the Comparison Test be used to prove absolute convergence?

Yes, the Comparison Test can be used to prove absolute convergence. If the terms of a series are less than or equal to the terms of an absolutely convergent series, then the original series is also absolutely convergent.

Replies
2
Views
697
Replies
3
Views
578
Replies
2
Views
1K
Replies
5
Views
2K
Replies
1
Views
912
Replies
4
Views
616
Replies
6
Views
798
Replies
1
Views
1K
Replies
1
Views
953
Replies
4
Views
911