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

  • Thread starter Thread starter appgolfer
  • Start date Start date
appgolfer
Messages
1
Reaction score
0
I know that \sum of tan(1/x^2) converges, but why?
 
Physics news on Phys.org


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).
 


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.
 


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.

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!
 


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!

Sure you can! If 0<a<b and 0<c<d, then \frac{a}{d}&lt;\frac{b}{c}.
In your example, that would be 1<1.5 and 0.25<0.5, then \frac{1}{0.5}&lt;\frac{1.5}{0.25}, which is perfectly true.

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


micromass said:
Sure you can! If 0<a<b and 0<c<d, then \frac{a}{d}&lt;\frac{b}{c}.
In your example, that would be 1<1.5 and 0.25<0.5, then \frac{1}{0.5}&lt;\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!
 

Similar threads

Back
Top