Does the Series Sum of sin(1/n^2) from n=1 to Infinity Converge?

[seto6]

Homework Statement

given this find if it converge or diverge.

\sum^{n=0}_{infinity} sin(\frac{1}{n^2})

sin (1/n^2) as n goes from zero to infinity.

Homework Equations

The Attempt at a Solution

i tried

sin(1/n^2) <= sin(1/n) ~ 1/n

by squeeze theorem

0 <= sin(1/n^2) <= 1/n

as the limit goes to infinity the 1/n goes to zero thus sin(1/n^2) goes to zero, does that mean it converge

 

[Mark44]

Homework Statement

given this find if it converge or diverge.

\sum^{n=0}_{infinity} sin(\frac{1}{n^2})

Your LaTeX code was almost right. Click on my version to see what I did.
\sum_{n=1}^{\infty} sin(\frac{1}{n^2})

BTW, the index n really should start at 1, not 0, since the expression being summed is undefined at n = 0.

sin (1/n^2) as n goes from zero to infinity.

Homework Equations

The Attempt at a Solution

i tried

sin(1/n^2) <= sin(1/n) ~ 1/n

by squeeze theorem

0 <= sin(1/n^2) <= 1/n

as the limit goes to infinity the 1/n goes to zero thus sin(1/n^2) goes to zero, does that mean it converge

In a series
\sum_{n=0}^{\infty} a_n
If
\lim_{n \to \infty} a_n~=~0
then you really don't know anything about the series. For example, lim a_n = 0 for both of the following series, but the first one diverges and the second converges.

\sum_{n=0}^{\infty} \frac{1}{n}

\sum_{n=0}^{\infty} \frac{1}{n^2}

 

[Mark44]

This was a good idea - sin(1/n) ~ 1/n
It's not right on the money, but it's a start.

The limit comparison test is good to know.

 

[seto6]

would it be valid to compare it with
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br />

since sine if always less than one (book is telling me its a non negative series)

and very close to zero (when n is large) its almost like
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br />

therefore
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br /> coverage thus sine should

 

[lanedance]

all your latex are the same?

sin(\frac{1}{n^2}) &lt; sin(\frac{1}{n})
for all n>1 seemed like a good start

then as sin(\frac{1}{n}) \approx \frac{1}{n} for large n can you think of anything similar in form to \frac{1}{n} but such that sin(\frac{1}{n}) is less than it for all n?

 

[Mark44]

would it be valid to compare it with
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br />

No, because you don't want to compare your series with itself. Why would you want to do that?

since sine if always less than one (book is telling me its a non negative series)

and very close to zero (when n is large) its almost like
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br />

therefore
<br /> \sum_{n=1}^{\infty} sin(\frac{1}{n^2})<br /> coverage thus sine should

?