Series

I have the suspicion that the following series converges, but how would I prove it? I'm guessing that for some n, $\sin(\frac{1}{n^2}) < \frac{1}{n^2}$ so, as the later converges, the former must, too. But how should I prove it?

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

Note the absolute value.

>>1
sin(1/n^2) < 1/n^2 for all positive integers n. First note that 1/1^2 = 1 > sin(1/1^2) ~= 0.8414. Now, the derivative of 1/n^2 is -2/n^3. The derivative of sin(1/n^2) is cos(1/n^2)*-2/n^3. If we look at n >= 1, then 1/n^2 is in the range (0,1], and cosine of that range is [cos(1),1). This means that for every possible value of n >= 1, cos(1/n^2) < 1, so the derivative of sin(1/n^2) is less than -2/n^3. So, the derivative of sin(1/n^2) is less than the derivative of 1/n^2 for all n > 1, and since 1/1^1 > sin(1/1^2), this allows us to conclude that sin(1/n^2) < 1/n^2 for n >= 1.

Thank you!

sin(1/n^2) < 1/n^2 for all positive integers n.

Lol what? sin(1/n^2) <= 1 for all positive integers n. Fixed.

umm...nevermind lol

there's actually a simpler way than >>3 gives. you can show sin(x) <= x for all positive x; the derivative of sin(x) is cos(x), which has absolute value <= 1, and the derivative of x is 1 everywhere. since sin(0) = 0, sin(x) <= x for positive x.

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