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Cambree's question at Yahoo! Answers (Convergence of a sequence)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. If it diverges to infinity, state your answer as "INF" (without the quotation marks). If it diverges to negative infinity, state your answer as "MINF". If it diverges without being infinity or negative infinity, state your answer as "DIV".

limit as n approaches infinity -> (29/19^(n))+ 18arctan(n^5)
Here is a link to the question:

Determine whether the sequence is divergent or convergent? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Cambree,

Easily proved, $29/19^n$ is bounded below (by $0$) and decreasing, so it is convergent. The sequence $18\arctan n^5$ is bounded above (by $18\pi/2$) and increassing, so it is convergent. As a consequence, the given sequence is convergent. Besides, $$\lim_{n\to +\infty}\left(\frac{29}{19^n}+ 18\arctan n^5\right)=\frac{19}{+\infty}+18\arctan (+\infty)=0+18\frac{\pi}{2}=\boxed{9\pi} $$ If you have further questions, you can post them in the Pre-Calculus section.