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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.

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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?

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  • $\begingroup$ 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

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