`sum_(n=0)^oo n!(x/2)^n` Find the values of x for which the series converges.

Expert Answers

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For the power series `sum_(n=0)^oo n!(x/2)^n,` we may apply Ratio Test.

In Ratio test, we determine the limit as:

`lim_(n-gtoo)|a_(n+1)/a_n| = L`


`lim_(n-gtoo)|a_(n+1)*1/a_n| = L`

 Then ,we follow the conditions:

a) `L lt1` then the series converges absolutely

b) `Lgt1` then the series diverges

c) `L=1 ` or does not exist  then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.

The given power series `sum_(n=0)^oo n!(x/2)^n` has:

`a_n =n!(x/2)^n`



       ` =1/(n!)(2^n/x^n)`

       ` =2^n/((n!)x^n)`

`a_(n+1) =(n+1)!(x/2)^(n+1)`

            ` = (n+1)(n!) x^(n+1)/2^(n+1)`

           ` = (n+1)(n!)(x^n*x)/(2^n*2)`

           ` =((n+1)(n!)*x^n*x)/(2^n*2))`

Applying the Ratio test on the power series, we set-up the limit as:

`lim_(n-gtoo) |((n+1)(n!)*x^n*x)/(2^n*2)*2^n/((n!)x^n)|`

Cancel out common factors: `x^n,` `n!` , and `2^n` .

`lim_(n-gtoo) |((n+1)x)/2|`

Evaluate the limit.

`lim_(n-gtoo) |((n+1)*x)/2| = |x/2|lim_(n-gtoo) |n+1|`

                            ` = |x/2|* oo`

                            ` = oo `       

The limit value `L= oo ` satisfies `Lgt 1` for all `x.`

Therefore,  the power series `sum_(n=0)^oo n!(x/2)^n`  diverges for all `x` .

There is no interval for convergence.

Note: The radius of convergence is `0` . The `x=0` satisfy the convergence at points where `n!(x/2)^n=0` .

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