Series 2

Also sprach Zarathustra

Is the next inf. series is convergence?

Sigma(1-->inf) 5^k*tan(pi/6^k)

I think it is, using Sigma(1-->inf) 5^k / 6^k < inf.
...but I can't prove it...

Thank you all!!!

dwsmith

MHF Hall of Honor
Is the next inf. series is convergence?

Sigma(1-->inf) 5^k*tan(pi/6^k)

I think it is, using Sigma(1-->inf) 5^k / 6^k < inf.
...but I can't prove it...

Thank you all!!!
Ratio test

Failure

Is the next inf. series is convergence?

Sigma(1-->inf) 5^k*tan(pi/6^k)

I think it is, using Sigma(1-->inf) 5^k / 6^k < inf.
...but I can't prove it...

Thank you all!!!

Consider that $$\displaystyle \tan(x)=x+o(x^2)$$, for $$\displaystyle x\to 0$$, and, therefore, you have that

$$\displaystyle 5^k\tan\frac{\pi}{6^k}=5^k\cdot \left(\frac{\pi}{6^k}+o\left(\big(\frac{\pi}{6^k}\big)^2\right)\right) = \left(\frac{5}{6}\right)^k\cdot\pi + o\left(\left(\frac{5}{6}\right)^k\right)$$, for $$\displaystyle k\to\infty$$.