Question

Question #7fb9f

Answer 1

`beta=e`

Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

`S=sum_(r=1)^oo a_n \ \`, and `\ \ L=lim_(n rarr oo) |a_(n+1)/a_n|`

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

Our series is;

`S = sum_(n=1)^oo a_k` with `a_n=(x^n n!)/n^n`

So our test limit is:

`L = lim_(n rarr oo) | (( x^(n+1)(n+1)!)/( (n+1)^(n+1) ) )/ (( x^n n!)/n^n ) |`
`\ \ \ = lim_(n rarr oo) | ( x^(n+1)(n+1)!)/( (n+1)^(n+1) ) * (n^n)/( x^n n!) |`
`\ \ \ = lim_(n rarr oo) | ( x x^n(n+1)n!)/( (n+1)(n+1)^n ) * (n^n)/( x^n n!) |`
`\ \ \ = lim_(n rarr oo) | ( x )/( (n+1)^n ) * n^n |`
`\ \ \ = |x| \ lim_(n rarr oo) | ( n/(n+1))^n |`
`\ \ \ = |x| \ 1/e`
`\ \ \ = |x/e|`

Comparing with the definition of the question, we can conclude that the series converges if `L lt 1`

`| x/e | lt 1 => |x| lt e`

Similarly, the series diverges if `L gt 1`

`| x/1e| gt 1 => |x| gt e`

Hence, `beta=e`

Answer 2

See below.

Explanation:

Using the asymptotic Stirling formula

`k! approx sqrt(2pi k)(k/e)^k` we have

`sum_(k=1)^oo (x^k k!)/k^k approx sum_(k=1)^oo sqrt(2pi k) (x/e)^k` and also

`sum_(k=1)^oo (x/e)^k le sum_(k=1)^oo sqrt(2pi k) (x/e)^k le esum_(k=1)^oo (k+1)/e(x/e)^k`

so we conclude that the convergence is for `abs(x/e) < 1` or `absx < e`