# Convergence of series with trigonometric function :(

• Feb 26th 2009, 11:25 PM
wyhwang7
Convergence of series with trigonometric function :(
hello, this is the problem:

show that

infinite
Σ sin (1/n^2) is a positive, convergent series.
n=1

Hint: use the inequality sinx is less than or equal to x for x is greater than or equal to 0
• Feb 26th 2009, 11:55 PM
TwistedOne151
p-series
Are you familiar with the convergence/divergence of the p-series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p}$?

--Kevin C.
• Feb 27th 2009, 12:35 AM
wyhwang7
yes, but can this be applied even though the exponent is within the trigonometric function?
• Feb 27th 2009, 12:55 AM
Jameson
Quote:

Originally Posted by wyhwang7
yes, but can this be applied even though the exponent is within the trigonometric function?

Not exactly. First show that 1/n^2 is a converging series. Then use the hint the problem gave, that for n>0, $\displaystyle \sin(n) \le n$. Put another way, $\displaystyle \frac{1}{\sin^2(n)} \le \frac{1}{n^2}$ So we have established a bounding series that converges and this proves that the other series converges as well.
• Feb 27th 2009, 01:07 AM
wyhwang7
thanks so much! but if http://www.mathhelpforum.com/math-he...7321dbc9-1.gif, wouldn't http://www.mathhelpforum.com/math-he...117f6adc-1.gif be the reverse? because n^2 would be the larger denominator
• Feb 27th 2009, 01:31 AM
Jameson
I made a typo. The sine should be around the whole expression, not just the denominator. The way I wrote it was wrong, sorry.