Also sprach Zarathustra Dec 2009 1,506 434 Russia May 14, 2010 #1 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!!!

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!!!

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 14, 2010 #2 Also sprach Zarathustra said: 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!!! Click to expand... Ratio test

Also sprach Zarathustra said: 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!!! Click to expand... Ratio test

Failure Jul 2009 555 298 Zürich May 15, 2010 #3 Also sprach Zarathustra said: 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!!! Click to expand... 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 \(\displaystyle k\to\infty\). Reactions: Also sprach Zarathustra

Also sprach Zarathustra said: 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!!! Click to expand... 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 \(\displaystyle k\to\infty\).