The discussion focuses on the application of the Ratio Test to determine the convergence of the series defined by the terms an = 2^n/n!. Participants confirm the correct approach by calculating the next term an+1 = 2^(n+1)/(n+1)! and simplifying the ratio |a_{n+1}/a_{n}|. The cancellation of factorial terms is highlighted as a key step in the process, demonstrating the effectiveness of the Ratio Test in this context.
PREREQUISITESStudents studying calculus, particularly those focusing on series and convergence tests, as well as educators looking for clear examples of the Ratio Test application.
NastyAccident said:Yes.
When in doubt, set a term an equal to what is being summed. In this case, an = 2^n/n!.
Now, get the next term in the series so an+1. In this case, an+1 = 2^(n+1)/(n+1)!.
Then, plug in an and an+1 into [tex]|\frac{a_{n+1}}{a_{n}}|[/tex]
Now, the n! cancels because n! is equivalent to n*(n-1)(n-2)(n-3)(n-4). Whereas (n+1)!, is equivalent to (n+1)(n)(n-1)(n-2)(n-3)(n-4).
So, as you can see, the terms to the right of (n+1) are canceled out.
NastyAccident