Convergence of series with trigonometric function :(

**wyhwang7** · February 26th 2009, 11:25 PM

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

**TwistedOne151** · February 26th 2009, 11:55 PM

Are you familiar with the convergence/divergence of the p-series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ ?

--Kevin C.

**wyhwang7** · February 27th 2009, 12:35 AM

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

**Jameson** · February 27th 2009, 12:55 AM

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, $\sin(n) \le n$ . Put another way, $\frac{1}{\sin^2(n)} \le \frac{1}{n^2}$ So we have established a bounding series that converges and this proves that the other series converges as well.

**wyhwang7** · February 27th 2009, 01:07 AM

thanks so much! but if

, wouldn't

be the reverse? because n^2 would be the larger denominator

**Jameson** · February 27th 2009, 01:31 AM

I made a typo. The sine should be around the whole expression, not just the denominator. The way I wrote it was wrong, sorry.

Thread: Convergence of series with trigonometric function :(

LinkBack

Thread Tools

Display

Convergence of series with trigonometric function :(

p-series

Similar Math Help Forum Discussions

Convergence of the Taylor series for the sine function

radius of convergence and interval of convergence of the series

Testing for Convergence of (Trigonometric) Series

Function series pointwise convergence question

Proove that convergence of one series implies the convergence of another series

Search Tags