Question

Consider the power series \[ \text { (i) } \sum \frac{n !}{n^{n}} x^{n} \quad \text { and } \quad \text { (ii) } \sum \frac{n^{n}}{n !} x^{n} . \] a) Show that their radii of convergence are $ e $ and $ e^{-1} $ respectively. b) Show, by any method, that $ a(n)=\left(1+\frac{1}{n}\right)^{n} $ is a strictly increasing sequence (whose limit is $ e $ ). c) Deduce from your working in (a) and (b) that (i) diverges when $ x=\pm e $. d) Here you may assume Stirling's formula: \[ n ! /\left\{\sqrt{2 \pi} n^{n+1 / 2} e^{-n}\right\} \rightarrow 1 \quad \text { as } n \rightarrow \infty \] Show that series (ii) diverges when $ x=e^{-1} $. Using Stirling's formula and (b), show that (ii) is conditionally convergent when $ x=-e^{-1} $.

          Consider the power series
\[
\text { (i) } \sum \frac{n !}{n^{n}} x^{n} \quad \text { and } \quad \text { (ii) } \sum \frac{n^{n}}{n !} x^{n} .
\]
a) Show that their radii of convergence are \( e \) and \( e^{-1} \) respectively.
b) Show, by any method, that \( a(n)=\left(1+\frac{1}{n}\right)^{n} \) is a strictly increasing sequence (whose limit is \( e \) ).
c) Deduce from your working in (a) and (b) that (i) diverges when \( x=\pm e \).
d) Here you may assume Stirling's formula:
\[
n ! /\left\{\sqrt{2 \pi} n^{n+1 / 2} e^{-n}\right\} \rightarrow 1 \quad \text { as } n \rightarrow \infty
\]
Show that series (ii) diverges when \( x=e^{-1} \). Using Stirling's formula and (b), show that (ii) is conditionally convergent when \( x=-e^{-1} \).

$Consider the power series (i) ∑(n !)/(n^n) x^n and (ii) ∑(n^n)/(n !) x^n . a) Show that their radii of convergence are e and e^-1 respectively. b) Show, by any method, that a(n)=(1+(1)/(n))^n is a strictly increasing sequence (whose limit is e ). c) Deduce from your working in (a) and (b) that (i) diverges when x=± e. d) Here you may assume Stirling's formula: n ! /{√(2 π) n^n+1 / 2 e^-n}→ 1 as n →∞ Show that series (ii) diverges when x=e^-1. Using Stirling's formula and (b), show that (ii) is conditionally convergent when x=-e^-1.$

Added by Ahsan N.

Question

Please give Ace some feedback