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Find the disk of convergence for each of the following complex power series.

∑n=0∞(n!)2zn(2n)!

Short Answer

Expert verified

Hence, the required disk of convergence is.

|z|<4

Step by step solution

01

Define Disk of Convergence 

For any power series∑anzn where z is a complex numbers, then disk of convergence is given by: .ÒÏ=limn→∞|z×nn+1|=|z|

02

Step 2:Find the disk of Convergence 

The given power series is:∑n=0∞(n!)2zn(2n)!, where, .an=(n!)2zn(2n)!

Now, let us evaluate the ratio as:

ÒÏ=limn→∞|an+1an|=limn→∞|((n+1)!)2zn+1(2(n+1))!(n!)2zn(2n)!|=limn→∞|zâ‹…(n+1)2(n!)2(n!)2â‹…(2n)!2(n+1)(2n+1)(2n)!|=limn→∞|z2â‹…(n+1)2(n+1)(2n+1)|

On further evaluating as:

ÒÏ=limn→∞|z2â‹…(n+1)2(n+1)(2n+1)|=|z|2limn→∞|(n+1)(2n+1)|=|z|2limn→∞|(1+1n)(2+1n)|=|z|2|(1+0)(2+0)|=|z|4

Now, for the series to be convergent, we have .ÒÏ<1 So,

ÒÏ<1|z|4<1|z|<4

Hence, the required disk of convergence is.|z|<4

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