 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-ii/3/b/ii/solution

(n!)^{1/ n} \sim \frac{n}{e} (2 πn)^{1/ (2 n)}, ∣ a_{n} ∣^{1/ n} = \frac{n ^{1 + 1/ n^{2}}}{( n ! ) ^{1/ n}} ⟶ e .

Here

(2 πn)^{1/ (2 n)} \to 1

and

n^{1/ n^{2}} \to 1

R = e^{- 1}

 New to topics? Read the docs here!