0
$\begingroup$

Consider the power series $\sum a_n x^n$ where

$$ a_n = {k \choose n} $$

for some $k$. What is the radius of convergence of this power series? I got one. Does that seem correct?


I got that the radius of convergence is one as follows: In Ross' text, we define the radius of convergence as $1/\beta$ where $\beta = \lim\left|a_{n+1} / a_n \right|$. I just considered

$$ \lim\left|{k \choose n+1} \cdot {k \choose n}^{-1} \right| $$

and working out the algebra, got $\beta = 1$ so the radius of convergence is one.

$\endgroup$
1
$\begingroup$

When $n>k$, $a_n=0$. So the series is actually a polynomial.

$\endgroup$
  • $\begingroup$ So is the radius of convergence not one? I used the standard $\lim\left|a_{n+1}/a_n\right|$ to get one $\endgroup$ – Jordan Kelk Apr 9 '14 at 3:03
  • $\begingroup$ @JordanKelk If you show your work, we can correct it. $\endgroup$ – Pedro Tamaroff Apr 9 '14 at 3:05
  • $\begingroup$ I've added my work above $\endgroup$ – Jordan Kelk Apr 9 '14 at 3:08
  • 1
    $\begingroup$ @JordanKelk "...and working out the algebra." That is what you have to show us...! $\endgroup$ – Pedro Tamaroff Apr 9 '14 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.