appgolfer
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I know that \sum of tan(1/x^2) converges, but why?
micromass said:For a more rigourous explanation, apply the inequalities
\sin\left(\frac{1}{x^2}\right)\leq \frac{1}{x^2}~\text{and}~1-\frac{1}{2x^4}\leq \cos\left(\frac{1}{x^2}\right)
These inequalities follow immediately from regarding the Taylor series of sin and cos. This also gives you an inequality:
\tan\left(\frac{1}{x^2}\right)\leq\frac{\frac{1}{x^2}}{1-\frac{1}{2x^4}}
Working this out should give you a nice upper bound of the tangent function. So the answer follows from the comparison test.
Mute said:You can't divide inequalities.
e.g., 1 < 1.5 and 0.25 < 0.5, but that doesn't mean that 1/0.25 < 1.5/0.5 => 4 < 3!
micromass said:Sure you can! If 0<a<b and 0<c<d, then \frac{a}{d}<\frac{b}{c}.
In your example, that would be 1<1.5 and 0.25<0.5, then \frac{1}{0.5}<\frac{1.5}{0.25}, which is perfectly true.
You have to be careful WHICH PART of the inequality you divide though...