The discussion centers on finding the radius and interval of convergence for the series defined as (n(x-4)^n) / (n^3 + 1) from n=1 to infinity. Participants utilized the Ratio Test, leading to the conclusion that the radius of convergence is 1, resulting in the interval (3, 5). The need for careful evaluation of limits and the importance of absolute values in convergence tests were emphasized, particularly when determining behavior at the endpoints x=3 and x=5.
PREREQUISITESStudents studying calculus, particularly those focusing on series and convergence, as well as educators seeking to clarify concepts related to series convergence tests.
Now, since n2 + 1/n > n2, for all n >= 1,arl146 said:ok well
we need \frac{n}{n^3+1} < \frac{1}{n^2} for the test
and that can be proven by:
\frac{n}{n^3+1} = \frac{n(1)}{n(n^2+1/n)} = \frac{1}{n^2+1/n}
arl146 said:so now we are looking at \frac{1}{n^2+1/n} < \frac{1}{n^2} for my series to be convergent since \frac{1}{n^2} converges.
looking at the denominators: n2+\frac{1}{n} is > n2
thus making \frac{n}{n^3+1} = \frac{1}{n^2+1/n} < \frac{1}{n^2} this true
is that good enough?