Does the series (sin(1/n))/sqrt(n) converge or diverge?

[lax1113]

Infinite series sin(1/n)/n ?

Homework Statement

does the series (sin (1/n)) / sqrt ( n ) converge or diverge? (series from n = 1 to infinity...)

Homework Equations

The Attempt at a Solution

I thought that for this we could do a comparison of sin (1/n) to a finite number. Let's say we use the number -2. -2 is less than sin(1/n) for all n. With that being said, -2/ sqrt( n ) diverges, as it is a p series with p less than 1.
But, this series actually converges... what did I do wrong?

 

[hgfalling]

Suppose instead of comparing sin(1/n) to a finite number, you compared it to a function. As n->infinity, what do you know about sin(1/n) and its relationship to some function that might be easier to work with?

(By the way, the term comparison tests only work if the terms of both series are all positive. That's why you can't use -2 as a comparison term to show that your series diverges.)

 

[sutupidmath]

You could probbably use the fact that

[tex]|sin\theta|<\theta[/tex]

Edit: I think this is basically what hgfalling was suggesting. That is [tex]sin(\theta)=O(\theta), \mbox{ or } sin(\theta)=\Omega (\theta) \mbox{ as } \theta\rightarrow 0. [/tex]

 

[Sweet_GirL]

You can use the comparison tests
since sin(1/n) is positive since the angle (1/n) is in the first quadratic for n=1,2,3,...

To test it, you could use the limit comparison test with a p-series, can you do that ?

 

[Gregg]

Show it converges faster than 1/x?

 

[dacruick]

as n --> infinity, 1/n ---> 0. sin(0) = 0. You can literally say that because the value at infinity is 0, it converges.