# Trig. integrals

1. Apr 28, 2008

### RyanMcStylin

1. The problem statement, all variables and given/known data
So this problem is actually an infinite series question, but i have shaved it down to an integral that i cannot seem to solve. I am new to this so i am going to try to talk you through it
here is the original problem:
the sum from n=1 to n=$$\infty$$ of sin(1/n)

2. Relevant equations
we have learned multiple ways to solve these equations, for this i have decided to use the integral test which says, if the integral from 1 to $$\infty$$ converges, then the sum converges.

3. The attempt at a solution

so far i have simplified it down to:
the integral from 1 to $$\infty$$ sin(1/x)dx however 1/x is also equal to x^-1 and taking this anti-derivative leaves me with x all over the place and to many options for an anti-derivative. i was thinking u substitution, any ideas.

2. Apr 28, 2008

### Dick

In fact, you don't have any options for an antiderivative. You can't integrate that in a simple form. Forget the integral test. Think about a comparison test and the limiting behavior of sin(x) as x->0 which is the same thing as sin(1/n) as n->infinity.

3. Apr 30, 2008

### RyanMcStylin

good plan, but i figured it out. in case you were wondering. since 1/n is always such a small angle the entire sine part of the equation is negligible. so when determining convergence of an infinite series involving sine, determine the convergence of the angle and you have the convergence of the whole function. in this case since 1/n diverges by p-series so does sin(1/n). thanks for the help

4. Apr 30, 2008

### frumdogg

I have learned it is best to throw out the portion of the equation with the n in the denominator. Makes it less messier.