Power Series and the Radius of Convergence

In Section 5.6, we met the idea of writing f (x) = P_n(x) + R_n(x), to express a function in terms of its Taylor polynomial, together with a remainder. We even saw in 5.30 that, for some functions, the remainder R_n(x)$\to$ 0 as n$\to$$\infty$ for each fixed x. We now recognise this as showing that certain series converge.

We have more effective ways of showing that such a series converges -- we can use that ratio test. But note that such a test will only show that a series converges, not that it converges to the function used to generate it in the first place. We saw an example of such a problem in the Warning before Example 5.36.

To summarise the results we had in Section 5.6,

Proposition 7.1   The following series converge for all values of x to the functions shown:

$$\nolimits^{x}_{} {\frac{{x^2}}{{2!}}} {\frac{{x^2}}{{2!}}} {\frac{{x^n}}{{n!}}}$$

$${\frac{{x^3}}{{3!}}} {\frac{{x^5}}{{5!}}} {\frac{{x^{2n+1}}}{{(2n+1)!}}}$$ sin x

$${\frac{{x^2}}{{2!}}} {\frac{{x^4}}{{4!}}} {\frac{{x^{2n}}}{{2n!}}}$$ cos x

$${\frac{{x^3}}{{3!}}} {\frac{{x^5}}{{5!}}} {\frac{{x^{2n+1}}}{{(2n+1)!}}}$$ sinh x

$${\frac{{x^2}}{{2!}}} {\frac{{x^4}}{{4!}}} {\frac{{x^{2n}}}{{2n!}}}$$ cosh x

These are all examples of the subject of this section; they are real power series, which we can use to define functions. The corresponding functions are the best behaved of all the classes of functions we meet in this course; indeed are as well behaved as could possibly be expected. We shall see in this section that this class of functions are really just ``grown up polynomials'', and that almost any manipulation valid for polynomials remains valid for this larger class of function.

Definition 7.2   A real power series is a series of the form $\sum$a_nxⁿ, where the a_n are real numbers, and x is a real variable.

We are thus dealing with a whole collection of series, one for each different value of x. Our hope is that there is some coherence; that the behaviour of series for different values of x are related in some sensible way.

Example 7.3   The geometric series $$\sum_{{n=0}}^{\infty}$$xⁿ is another example of a power series we have already met. We saw this series is convergent for all x with | x| < 1.

It turns out that a power series is usually best investigated using the ratio test, Theorem 6.21. And the behaviour of power series is in fact very coherent.

Theorem 7.4 (Radius of Convergence)   Suppose $\sum$a_nxⁿ is a power series. Then one of the following happens:

• $\sum$a_nxⁿ converges only when x = 0; or
• $\sum$a_nxⁿ converges absolutely for all x; or
• there is some number R > 0 such that $\sum$a_nxⁿ converges absolutely for all x with | x| < R, and diverges for all x with | x| > R.

No statement is made in the third case about what happens when x = R.

Definition 7.5   The number R described above is called the radius of convergence of the power series. By allowing R = 0 and R = $\infty$, we can consider every power series to have a radius of convergence.

Thus every power series has a radius of convergence. We sometimes call the interval (- R, R), where the power series is guaranteed to converge, the interval of convergence. It is characterised by the fact that the series converges (absolutely) inside this interval and diverges outside the interval.

• The word ``radius'' is used, because in fact the same result is true for complex series, and then we have a genuine circle of convergence, with convergence for all (complex) z with | z| < R, and guaranteed divergence whenever | z| > R.

• Note the power of the result; we are guaranteed that when | x| > R, the series diverges; it can't even converge ``accidentally'' for a few x's with | x| > R. Only on the circle of convergence is there ambiguity.

This regularity of behaviour makes it easy to investigate the radius of convergence of a power series using the ratio test.

Example 7.6   Find the radius of convergence of the series $$\sum$$$${\frac{{n x^n}}{{2^{n+1}}}}$$.

Solution. Recall that the ratio test only applies to series of positive terms, so we look at the ratio of the moduli.

$${\frac{{\vert\mbox{$(n+1)^{\rm st}$\ term}\vert}}{{\vert\mbox{$n^{\rm th}$\ term}\vert}}} {\frac{{(n+1)\vert x\vert^{n+1}}}{{2^{n+2}}}} {\frac{{2^{n+1}}}{{n\vert x\vert^n}}}$$ =

$${\frac{{(n+1)\vert x\vert}}{{n.2}}} \to {\frac{{\vert x\vert}}{{2}}} \to \infty$$ =

Thus the given series diverges if | x| > 2 and converges absolutely (and so of course converges) if | x| < 2. Hence it has radius of convergence 2.

Example 7.7   Find the radius of convergence of the series $$\sum$$$${\frac{{(-1)^n n! x^n}}{{n^n}}}$$.

Solution. This one is a little more subtle than it looks, although we have met the limit before. Again we look at the ratio of the moduli of adjacent terms.

$${\frac{{\vert\mbox{$(n+1)^{\rm st}$\ term}\vert}}{{\vert\mbox{$n^{\rm th}$\ term}\vert}}} {\frac{{ (n+1)! \vert x\vert^{n+1}}}{{(n+1)^{n+1}}}} {\frac{{n^n}}{{n! \vert x\vert^n}}} {\frac{{n^n}}{{(n+1)^{(n+1)}}}}$$ =

$${\frac{{n^n}}{{(n+1)^n}}} \left(\vphantom{\frac{n}{n+1}}\right. {\frac{{n}}{{n+1}}} \left.\vphantom{\frac{n}{n+1}}\right)^{n}_{}$$ =

$${\frac{{\vert x\vert}}{{\displaystyle \left(\frac{n+1}{n}\right)^n}}} \to {\frac{{\vert x\vert}}{{\mathop\mathrm{e}\nolimits }}} \to \infty$$ =

Here we have of course used the result about e$\nolimits$ given in Section 3.1 to note that

$$\left(\vphantom{1 + \frac{1}{n}}\right.$$1 + $${\frac{{1}}{{n}}}$$$$\left.\vphantom{1 + \frac{1}{n}}\right)^{n}_{}$$$$\to$$e$$\nolimits$$    as n$\to$$\infty$.    

Thus the given series diverges if | x| > e$\nolimits$ and converges absolutely (and so of course converges) if | x| < e$\nolimits$. Hence it has radius of convergence e$\nolimits$.

Exercise 7.8   Find the radius of convergence of the series $$\sum_{{n=0}}^{\infty}$$$${\frac{{x^n}}{{n^2+1}}}$$

We noted that the theorem gives no information about what happens when x = R, i.e. on the circle of convergence. There is a good reason for this -- it is quite hard to predict what happens. Consider the following power series, all of which have radius of convergence 2.

$$\sum_{{n=1}}^{\infty}$$$${\frac{{x^n}}{{2^n}}}$$        $$\sum_{{n=1}}^{\infty}$$$${\frac{{x^n}}{{n2^n}}}$$        $$\sum_{{n=1}}^{\infty}$$$${\frac{{x^n}}{{n^2 2^n}}}$$.

The first is divergent when x = 2 and when x = - 2, the second converges when x = - 2, and diverges when x = 2, while the third converges both when x = 2 and when x = - 2. These results are all easy to check by direct substitution, and using Theorem 6.29.