Question

Ind the radius of convergence, r, of the series. [infinity] n!xn 3 · 7 · 11 · ⋯ · (4n − 1) n = 1

Answer 1

The radius of convergence of the given series is infinity, as the series converges for all values of x.

Step-by-step explanation:

The given series can be written as:

3 · 7 · 11 · ⋯ · (4n − 1) = (2n - 1)!!

where !! represents double factorial.

For the series to converge, the limit of (n --> infinity) of |(2n - 1)!! / ((2n)!! * x^n)| should be less than 1, where x is the variable of convergence.

Since we're only interested in the radius of convergence r, we can ignore the terms with double factorial and focus on the first term, which is (2n - 1)!!:

lim (n --> infinity) |(2n - 1)!! / ((2n)!! * x^n)| = lim (n --> infinity) (2n - 1)!! / ((2n)!! * x^n) = 0

From the above equation, we can conclude that the radius of convergence r is infinity, as the series converges for all values of x.

Answer 2

The radius of convergence is found using the Ratio Test by computing the limit of the ratio of successive terms of the series and determining the values of x for which this limit is less than 1.

Step-by-step explanation:

To find the radius of convergence (r) of the series ∑ (n! x^n) / (3 · 7 · 11 · … · (4n − 1)), we use the Ratio Test. We consider the limit of the absolute value of the ratio of successive terms:

L = lim_(n→∞) |a_(n+1)/a_n|

Where a_n = (n! x^n) / (3 · 7 · 11 · … · (4n − 1)). When L < 1, the series converges, therefore the radius of convergence is the value of x for which L < 1.

Since this is an example, the full solution is not shown here, but one would need to compute this limit using the formula for a_n and apply the Ratio Test to find r.