i did a workshop recently about writing a series as a power series and then finding the radius of convergence, i'm perfectly happy finding the radius of convergence when it's in power series form $\sum_{n=0}^{\infty} a_{n}(z-a)^{n}$ however i was given two questions that looked like a power series however starting from n≥7
$\sum_{n≥7}^{\infty} (3i-1)^{n}(z-i)^{7n-1}$
and starting from n≥1000
$\sum_{n≥1000}^{\infty} (i)^{n}\frac{z^{2n-1}}{n!}$
And i have no clue how to put these into the usual power series form where i can then find the radius of convergence
There is no problem even whatever terms finite terms are removing as our series behaviour depend on large n
So we should follow usual procedure of finding radius of convergence
Only point is to considered that complex number does not have ordered so take absolute value
We apply Hadamard formula for finding ROC
$( |(3i-1)^{n}|^{1/n}|(z-i)|^{7}=\sqrt{10}|(z-i)|^7<1$ as $ |3i-1|=\sqrt{10}$
i.e $|z-i|<1/10^{7/2}$
SO ROC is $1/10^{7/2}$
Note: For series to converge $(a_n)^{1/n}<1$
Similarly
ROC of 2nd series is $\infty$
Try If You have problem Let me know