Series 2

Dec 2009
1,506
434
Russia
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!!!
 
Jul 2009
555
298
Zürich
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 \(\displaystyle k\to\infty\).