# Comparison Test Problem & Estimitaing it's error

• JRangel42
I'll try it out for myself.In summary, the conversation revolved around estimating the error in a series by using the sum of the first 10 terms. The person was unsure about how to approach this and was initially trying to find the integral, but it was pointed out that this was not necessary. Instead, they could use a simpler estimate to bound the error.

## Homework Statement

The question in the book is: "Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

My problem is estimating the error I'm looking for. I just need help with finding the integral.

## Homework Equations

((sin n)^2)/(n^3)

## The Attempt at a Solution

∞∫n ((sin n)^2)/(n^3)

u = ((sin n)^2) du = 2 ((sin n)^2)cos n dv = n^3 dx v = 1/4 x^4

This is as far as I got.

JRangel42 said:

## Homework Statement

The question in the book is: "Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

My problem is estimating the error I'm looking for. I just need help with finding the integral.

## Homework Equations

((sin n)^2)/(n^3)

## The Attempt at a Solution

∞∫n ((sin n)^2)/(n^3)

u = ((sin n)^2) du = 2 ((sin n)^2)cos n dv = n^3 dx v = 1/4 x^4

This is as far as I got.

What integral? Aren't you just given an infinite series, and are asked to estimate the error when you stop at 10 terms? Anyway, if you did have to integrate the function you would need to use the non-elementary function Ci(x). Maple 11 gets:
f:=sin(x)^2/x^3;
> Jt:=Int(f,x=1..t);
$$\int_1^\infty \frac{\sin^2(x)}{x^3} \, dx$$
> value(Jt) assuming t>1;
$$\frac{1}{4} \frac{t-t^2 \cos(2)+2t^2 \sin(2)-2\mbox{Ci}(2) t^2 - 2 + 2\cos^2(t) - 4t \cos(t) \sin(t) + 4\mbox{Ci}(2t) t^2}{t^2}$$

RGV

Last edited:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.

JRangel42 said:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.

Your error is the sum of (sin(k))^2/k^3 over all of the terms you didn't include. Don't sweat the Ci function. All you need is an estimate to bound the error. (sin(k))^2/k^3<=1/k^3. Estimate that instead.

JRangel42 said:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.

You still have not answered my question: what integral? Are you not just dealing with a sum? (The Ci stuff was just to show you that the integral is not 'doable', so you should not waste time on it!)

RGV

Alright guys. Thanks for the tip. I think get it now.