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

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- Feb 26th 2009, 11:25 PMwyhwang7Convergence 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 PMTwistedOne151p-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 AMwyhwang7
yes, but can this be applied even though the exponent is within the trigonometric function?

- Feb 27th 2009, 12:55 AMJameson
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 AMwyhwang7
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 AMJameson
I made a typo. The sine should be around the whole expression, not just the denominator. The way I wrote it was wrong, sorry.