# series - converge or diverge

• Nov 10th 2009, 02:56 AM
ArTiCK
series - converge or diverge
Hi all,

I am trying to determine if the following series converges or diverges:

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

Not sure if what i did is correct:

-1 <= sin(n) <= 1

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

1/n^2 <= [sin^2(n)]/ n^2 <= 1/n^2

Note: sin^2(n) = [sin(n)]^2

I then compared the series to 1/n^2 and determined that the series 1/n^2 is a harmonic p series, where p > 1 and so it converges. Thus, the series $
\sum_{n=1}^{\infty}\frac{sin^2(n)}{n^2}
$
converges.

Could someone tell me if i am doing this correctly

ArTiCk
• Nov 10th 2009, 02:59 AM
HallsofIvy
Quote:

Originally Posted by ArTiCK
Hi all,

I am trying to determine if the following series converges or diverges:

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

Not sure if what i did is correct:

-1 <= sin(n) <= 1

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

You mean -1<= sin^2(n)<= 1 but the comparison test uses absolute values that doesn't matter.

Quote:

1/n^2 <= [sin^2(n)]/ n^2 <= 1/n^2

Note: sin^2(n) = [sin(n)]^2

I then compared the series to 1/n^2 and determined that the series 1/n^2 is a harmonic p series, where p > 1 and so it converges. Thus, the series $
\sum_{n=1}^{\infty}\frac{sin^2(n)}{n^2}
$
converges.

Could someone tell me if i am doing this correctly