I know that [tex]\sum[/tex] of tan(1/x^2) converges, but why?

[appgolfer]

I know that $$\sum$$ of tan(1/x^2) converges, but why?

 

[mathman]

I presume x is supposed to be an integer, but you should specify domain.
In any case, for large x, tan(1/x²) ~ 1/x², which implies convergence (assuming x starts > 0).

 

[micromass]

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]

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.

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]

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!

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...

 

[Mute]

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.

I stand corrected!

You have to be careful WHICH PART of the inequality you divide though...

I guess I wasn't careful enough!