Consider the power series $\sum_{n=0}^{\infty}p(n)a_nz^n$. Let the degree of $p(x)=d$. Given that $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=R> 0$, find the radius of convergence of given power series.

My view :take $p(x)=a_1x^d+a_2x^{d-1}+\cdots+a_{d+1}$ Multiplying $p(n)$ and $a_n$, I get coefficients of power series. I apply ratio test using given limit. Since power add on multiplication, it should turn out that radius is $R+d or Rd$. Is this approach OK? Am I justified in taking $p(x)$ to be an expression like the way I have defined? Any other methods would be equally appreciated.


It's perfectly valid to take $p(x) = b_1x^d + b_2x^{d-1} + \cdots + b_{d+1}$ (although it is more common to write $p(x) = b_0 + b_1 x + \cdots + b_{d}x^d$. Also note that I have used $b$'s instead of $a$'s to avoid confusion between the polynomial coefficients and the terms $a_n$).

Then, applying the ratio test, we have

$$\lim_{n\to\infty} \frac{p(n+1) a_{n+1}z^{n+1}}{p(n)a_nz^n} = z\lim_{n\to\infty} \frac{p(n+1)}{p(n)} \lim_{n\to\infty} \frac{a_{n+1}}{a_n}.$$ The last limit we know is $R$ from the information given. It is also true that $$\lim_{n\to\infty} \frac{p(n+1)}{p(n)} = 1$$ as you can verify by substituting in your expression for $p(x)$ and dividing the top and bottom by $n^d$.

Now, can you solve for the radius of convergence from here?

  • $\begingroup$ Seems something wrong. This way I get 1/R as radius of convergence. $\endgroup$ – low iq Jun 2 '16 at 2:20
  • $\begingroup$ No, that's correct. The radius is still $1/R$. Since polynomials grow slower than any exponential, it makes sense that multiplying every term in a power series by a polynomial in $n$ does not change the radius of convergence, which is governed by the exponential term $z^n$. This is what you've shown. $\endgroup$ – Strants Jun 2 '16 at 3:52

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.