I have a series $S$ with general terms $a_n=\frac{(-1)^n(x-1)^n}{(2n-1)2^n}$, $n\ge 1$:

$$S = \sum_{n=1}^\infty \frac{(-1)^n(x-1)^n}{(2n-1)2^n}$$

Finding the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ and then finding the limit of the ratio as $n\to\infty$, I find the limit to be $1$ and the interval to be $-1 \lt x \lt 3$. More declaratively, the interval is $\left|\frac{x−1}{2}\right| \lt 1$ which I've refined to what was said earlier.

I've read conflicting sites that state the radius $R$ of convergence is $\frac{1}{N}$, where $N$ is the limit as found earlier, but also that it's half the interval length.

Here's my work:

$$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{\frac{(-1)^{n+1}(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\frac{(-1)^{n}(x-1)^{n}}{(2n-1)2^{n}}}\right| \\ &= \left|\frac{(-1)^{n+1}(x-1)^{n+1}(2n-1)(2^n)}{(-1)^n(x-1)^n(2(n+1)-1)(2^{n+1})}\right| \\ &= \left|\frac{(-1)(x-1)(2n-1)}{(2n+2-1)(2)}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+2}\right| \end{align}$$

Then, finding the limit $L$:

$$\begin{align} L &= \lim_{n\to\infty} \left(\left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+1}\right|\right) \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2n-1}{2n+1}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{\frac{2n}{n}-\frac{1}{n}}{\frac{2n}{n}+\frac{1}{n}}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2-\frac{1}{n}}{2+\frac{1}{n}}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{\lim_{n\to\infty} \left(2-\frac{1}{n}\right)}{\lim_{n\to\infty} \left(2+\frac{1}{n}\right)}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2}{2}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times 1 \\ &= \left|\frac{-(x-1)}{2}\right| \end{align}$$

Then I know my interval is $\left|\frac{-(x-1)}{2}\right| \lt 1$:

$$\left|\frac{-(x-1)}{2}\right| \lt 1 \\ -1 \lt \frac{x-1}{2} \lt 1 \\ -2 \lt x-1 \lt 2 \\ -1 \lt x \lt 3$$

If the limit found earlier is $1$, the radius would be $R = \frac{1}{1} = 1$, yet I've found the interval to be $(-1, 3)$, which would imply $R = 2$. Where have I made an error?

  • $\begingroup$ You should keep in mind that the power series is really in $z=(x-1)/2$. So with $R=1$, $-1<z<1$ becomes $-1<x<3$. $\endgroup$ – Chrystomath Jul 7 at 16:32

For a power series $$ \sum_{n=0}^\infty c_n (z-a)^n, $$ the radius of convergence is $R = \frac1N,$ where $$ N = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| $$ provided that the limit exists and is a real number. Other sources say simply that the radius is $$ R = \lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|, $$ which is equivalent except (arguably) in the case $N=0.$ See Ratio test and the radius of convergence.

Note that $c_n$ is not a term of the series; it's only a coefficient of a term of the series. The $n$th term is $a_n = c_n(z-a)^n.$

If you're looking at a site that says the radius of convergence is $\frac1N,$ this is the way they are most likely applying the ratio test. (Another possibility is that you have found a page with misinformation. Such things do exist on the web!)

You have defined $$a_n=\frac{(-1)^n(x-1)^n}{(2n-1)2^n},$$ so $a_n$ is not $c_n$ in the expression above. Instead, $a_n$ is a function of $x$ and the limit $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| $$ depends on the value of $x$ at which you evaluate it, as you showed in your calculations (which are correct). That's not the limit of ratios from which people derive the radius of convergence on pages like the ones you described. It would be nonsense for the radius of convergence to be a function of $x.$

The use of the limit $$ N = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| $$ to find the radius of convergence actually is based on the general ratio test that is defined for a general series. Namely, if you have a power series whose $n$th term is $a_n = c_n(x-a)^n,$ then $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| \lvert x-a\rvert = N \lvert x-a\rvert, $$ where $N \geq 0,$ provided that the limits exist and are real numbers. We have convergence by the general ratio test when $N \lvert x - a\rvert < 1,$ which (if $N > 0$) is true exactly when $$ \lvert x - a\rvert < \frac1N. $$

If we take the limit $N$ in the way it is meant to be taken on one of those "$\frac1N$" pages, we have $$c_n=\frac{(-1)^n}{(2n-1)2^n}$$ (note: everything that is in $a_n$ except the factor $(x-1)^n$) and therefore $$ N = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| = \frac12, $$ and the radius of convergence is $\frac1N = 2.$

This actually agrees with your calculations. You found that the limit of your ratio of terms was $\left|\frac{-(x-1)}{2}\right|.$ The thing is, $N$ is supposed to be multiplied by $|x-a|,$ not by $\left|\frac{-(x-a)}{2}\right|.$ But if you see that $$ \left|\frac{-(x-1)}{2}\right| = |x-1|\times\frac12 $$ then that factor $\frac12$ is your $N.$

Alternatively, we can compute $$ R = \lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to\infty}\left|\frac{\left(\frac{(-1)^n}{(2n-1)2^n}\right)} {\left(\frac{(-1)^{n+1}}{(2(n+1)-1)2^{n+1}}\right)}\right| = \lim_{n\to\infty}\left|\frac{-2(2n+1)}{2n-1}\right| = 2 $$ in order to get the radius of convergence $R.$

And then, since $a = 1,$ indeed the interval of convergence is $$\left(a - \frac{1}{N}, a + \frac{1}{N}\right) = (a - R, a + R) = (1 - 2, 1 + 2) = (-1, 3).$$

You're correct in your own calculations, but you're comparing them to a different set of calculations that are done in a slightly different way, even though they are justified by the same theorem and produce the same interval of convergence.

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  • $\begingroup$ "That's not the ratio test for a power series--or at least, it's not the ratio test from which people derive the radius of convergence." Looking at whether it's <1 is the ratio test. Finding the values of x such that that is less than 1 gives the convergence interval. $\endgroup$ – Acccumulation Jul 8 at 1:03
  • $\begingroup$ @Acccumulation Good point, it may be better to reserve the term "The Ratio Test" for the test that applies to general series, not just to coefficients of power series. The general Ratio Test and the ratio technique that I cite at the beginning of the answer are actually equivalent, but one requires you to take additional steps to work out the $x$ for which the absolute ratio is $<1$ while the other gives you the radius directly. $\endgroup$ – David K Jul 8 at 2:15
  • $\begingroup$ I see. If I only moved the $-\left|x-1\right|$ and not the denominator outside of the limit, I would have got $\frac{2}{4} = 2$ as the limit which runs congruent with expected outcome. It was just one superfluous step I did when I shouldn't have. $\endgroup$ – gator Jul 9 at 16:35

Note that a power series takes the form $$\sum_{n=0}^\infty a_n(x-x_0)^n$$ In your case you have $$a_n=\begin{cases}\frac{(-1)^n}{(2n-1)2^n}&n\ne0\\0&n=0\end{cases}\qquad x_0=1$$ If you calculate the limit you call $N$ we get $$N=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac12$$ So the radius of convergence is $R=1/N=2$ and hence the interval of convergence is $$x\in(x_0-R,x_0+R)=(-1,3)$$ as expected.

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  • $\begingroup$ I've edited my question to include my work. $N = \frac{1}{1} = 1$, not $\frac{1}{2}$? $\endgroup$ – gator Jul 7 at 16:57
  • $\begingroup$ @gator You have mistakenly included the term $(x-x_0)$ in the $a_n$ coefficients. You shouldn't get $(x-1)/2$ but rather $1/2$. $\endgroup$ – Peter Foreman Jul 7 at 16:59

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