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: Find the disk of convergence for each of the followingcomplex power series.

∑n=0∞(−1)nz2n(2n)!

Short Answer

Expert verified

The required disk of convergence is .|z|<∞

Step by step solution

01

Disk of Convergence 

For any power series ∑anznwhere 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∞(−1)nz2n(2n)!, wherean=(−1)nz2n(2n)!

Now, let us evaluate the ratio as:

ÒÏ=limn→∞|an+1an|=limn→∞|(−1)n+1z2(n+1)(2(n+1))!(−1)nz2n(2n)!|=limn→∞|(−1)z2(2n+1)(2n+2)|

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

ÒÏ=limn→∞|(−1)z2(2n+1)(2n+2)|<1|z|2<limn→∞|(2n+1)(2n+2)||z|2<∞|z|<∞

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

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