# Write … as a power series

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

• You could try writing out the first few terms of the series and see whether they can be written in terms of $n=0,1,2,3,\ldots$. – Jam Dec 1 '18 at 13:03
• 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. – 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

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

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