Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.

For example, find radius of convergence of these....

$\sum_{n=0}^{\infty}(-1)^n (x+3)^n$

For this I would use ratio test for power series and get radius of convergence = 1, fine.

but for


Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?

Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?

  • 2
    $\begingroup$ $\sum_{n=0}^{\infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html]. $\endgroup$ May 15, 2013 at 12:42
  • $\begingroup$ but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though... $\endgroup$
    – Tom
    May 15, 2013 at 13:02
  • $\begingroup$ when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one. $\endgroup$
    – GEdgar
    May 15, 2013 at 14:07

1 Answer 1


Let $$\sum _{n=0} ^{\infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:

$$\lim_{n\to \infty}{\Big|\frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}\Big|}<1$$ So,

$$\lim_{n\to \infty}{|(n + 1)(n + 2)x^2|}<1$$

Since for each $x\ne0$ we see $\lim_{n\to \infty}{|(n+1)(n+2)x^2|}=\infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.


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