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

  • $\begingroup$ You could try writing out the first few terms of the series and see whether they can be written in terms of $n=0,1,2,3,\ldots$. $\endgroup$ – Jam Dec 1 '18 at 13:03
  • $\begingroup$ Does this change anything if you set $a_i=0$ for $i\le n_0$ ? For the radius of convergence you are only interested in $a_n$ when $n\to\infty$ not the first terms. $\endgroup$ – zwim Dec 1 '18 at 13:07

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 $(a_n)^{1/n}<1$


ROC of 2nd series is $\infty$

Try If You have problem Let me know

  • $\begingroup$ isnt |3i-1|=\sqrt{10} $\endgroup$ – L G Dec 1 '18 at 14:27
  • $\begingroup$ Yes ,, You are right I had edited my answer'\ $\endgroup$ – SRJ Dec 1 '18 at 15:10

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