`sum_(n=0)^oo n!(x/2)^n` Find the values of x for which the series converges.
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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`
or
`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`
Then,
`1/a_n=1/(n!)(2/x)^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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