My textbook (Early Transcendentals 8th e., James Stewart) advises that in general to find the interval of convergence of a power series we should use the Ratio or Root Tests. However, I found that the interval of convergence could also be found by applying the Geometric Series Test: take the absolute value of the "common ratio", set it to less than $1$, and solve for $x$.

For example,

$$\sum_{n=1}^{\infty} \frac{x^n}{n^44^n} = \sum_{n=1}^{\infty} \frac{1}{n^4} (\frac{x}{4})^n$$

The "common ratio" is $r= \frac{x}{4}$ since it's the factor being raised to the power $n$.

A geometric series converges when $|r| < 1$

$$|\frac{x}{4}| < 1$$

$$-1 < \frac{x}{4} <1$$

$$-4 < x < 4$$

Which produces the same interval of convergence as when using the Ratio Test.

I found that this worked for the other power series presented in this section as well.

I'm aware that the power series in the example is NOT a geometric series because the coefficient of the series, $c_n = \frac{1}{n^4}$ is not constant as $n\rightarrow\infty$ and thus it does not actually have a common ratio since $r$ changes depending on which terms in the series are used to calculate it. In fact, none of the power series in this section were geometric series because none had a constant coefficient nor a true common ratio, hence why I'm unsure why the Geometric Series Test seemed to work for the power series presented.

Is using the Geometric Series Test to find the interval of convergence for power series valid? If so, why since not all power series are geometric series?


1 Answer 1


Even though you call it the "Geometric Series Test," the actual argument your proof describes is clearly the Ratio Test:

For example,

$\sum_{n=1}^∞ \frac{x^n}{n^4 4^n} = \sum_{n=1}^∞ \frac{1}{n^4} \left( \frac{x}{4} \right)$

The "common ratio" is $r = \frac{x}{4}$ since it's the factor being raised to the power $n$.

Here, $a_n = \frac{x^n}{n^4 4^n}$, so applying Ratio Test gives $$r = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \left( \frac{x}{4} \right) \lim_{n \to \infty} \frac{n^4}{(n+1)^4} = \frac{x}{4}.$$ The Ratio Test and the Root Test are both based on (and proven via) the condition for convergence of a geometric series. So it's not surprising that "pretending the series is geometric" works when the complicating factor is a rational function of $n$ like $\frac{1}{n^4}$, as that factor will multiply the limit by 1 in either the Ratio Test or the Root Test.

  • $\begingroup$ Given your explanation, is it valid to use the pseudo Geometric Series Test as a shortcut for evaluating power series instead of using the Ratio or Root Tests? $\endgroup$
    – Slecker
    Feb 10, 2021 at 5:32
  • 1
    $\begingroup$ It's a good heuristic if the coefficients in the series $\sum c_n x^n$ are of the form $c_n = \frac{p(n)}{q(n)} a^n$, where $p(n)$, $q(n)$ are polynomials. It might fail to determine convergence at the endpoints, though; and, I would write out the full Ratio/Root test correctly if I was answering a question on a HW or a test :) $\endgroup$ Feb 10, 2021 at 19:27
  • $\begingroup$ Thank you for the clarification! :) $\endgroup$
    – Slecker
    Feb 11, 2021 at 4:37

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