# 8.4: Boundary Issues and Abel’s Theorem

- Page ID
- 7976

- Explain the Abel's theorem

Summarizing our results, we see that any power series \(\sum a_nx^n\) has a radius of convergence \(r\) such that \(\sum a_nx^n\) converges absolutely when \(|x| < r\) and diverges when \(|x| > r\). Furthermore, the convergence is uniform on any closed interval \([-b,b] ⊂ (-r,r)\) which tells us that whatever the power series converges to must be a continuous function on \((-r,r)\). Lastly, if

\[f(x) = \sum_{n=0}^{\infty }a_nx^n\]

for \(x ∈ (-r,r)\), then

\[f'(x) = \sum_{n=0}^{\infty }a_nnx^{n-1}\]

for \(x ∈ (-r,r)\) and

\[\int_{t=0}^{x}f(t)dt = \sum_{n=0}^{\infty }a_n\frac{x^{n+1}}{n+1}\]

for \(x ∈ (-r,r)\).

Thus power series are very well behaved within their interval of convergence, and our cavalier approach from Chapter 2 is justified, EXCEPT for one issue. If you go back to Exercise Q1 of Chapter 2, you see that we used the geometric series to obtain the series,

\[\arctan x = \sum_{n=0}^{\infty } (-1)^n \frac{1}{2n+1}x^{2n+1}.\]

We substituted \(x = 1\) into this to obtain \(\frac{\pi }{4} = \sum_{n=0}^{\infty } (-1)^n \frac{1}{2n+1}\). Unfortunately, our integration was only guaranteed on a closed subinterval of the interval \((-1,1)\) where the convergence was uniform and we substituted in \(x = 1\). We “*danced on the boundary*” in other places as well, including when we said that

\[\frac{\pi }{4} = \int_{x=0}^{1}\sqrt{1 - x^2}dx = 1 + \sum_{n=1}^{\infty } \left ( \frac{\prod_{j=0}^{n-1}\left ( \frac{1}{2} - j \right )}{n!} \right ) \left (\frac{(-1)^n}{2n+1} \right )\]

The fact is that for a power series \(\sum a_nx^n\) with radius of convergence \(r\), we know what happens for \(x\) with \(|x| < r\) and \(x\) with \(|x| > r\). We never talked about what happens for \(x\) with \(|x| = r\). That is because there is no systematic approach to this boundary problem. For example, consider the three series

\[\sum_{n=0}^{\infty } x^n, \qquad \sum_{n=0}^{\infty } \frac{x^{n+1}}{n+1}, \qquad \sum_{n=0}^{\infty } \frac{x^{n+2}}{(n+1)(n+2)}\]

They are all related in that we started with the geometric series and integrated twice, thus they all have radius of convergence equal to \(1\). Their behavior on the boundary, i.e., when \(x = ±1\), is another story. The first series diverges when \(x = ±1\), the third series converges when \(x = ±1\). The second series converges when \(x = -1\) and diverges when \(x = 1\).

Even with the unpredictability of a power series at the endpoints of its interval of convergence, the *Weierstrass-M* test does give us some hope of uniform convergence.

Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum a_nr^n\) converges absolutely. Then \(\sum a_nx^n\) converges uniformly on \([-r,r]\).

**Hint**-
For \(|x| ≤ r\), \(|a_nx^n| ≤ |a_nr^n|\).

Unfortunately, this result doesn’t apply to the integrals we mentioned as the convergence at the endpoints is not absolute. Nonetheless, the integrations we performed in Chapter 2 are still legitimate. This is due to the following theorem by Abel which extends uniform convergence to the endpoints of the interval of convergence even if the convergence at an endpoint is only conditional. Abel did not use the term uniform convergence, as it hadn’t been defined yet, but the ideas involved are his.

Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum a_nr^n\) converges. Then \(\sum a_nx^n\) converges uniformly on \([0,r]\).

The proof of this is not intuitive, but involves a clever technique known as Abel’s Partial Summation Formula.

Let \(a_1, a_2, ..., a_n, b_1, b_2, ... , b_n\) be real numbers and let \(A_m = \sum_{k=1}^{m}a_k\). Then

\[a_1b_1 + a_2b_2 +···+ a_nb_n = \sum_{j=1}^{n-1}A_j(b_j - b_{j+1}) + A_nb_n\]

Prove Lemma \(\PageIndex{1}\).

**Hint**-
For \(j > 1\), \(a_j = A_j - A{j-1}\).

Let \(a_1, a_2, ..., a_n, b_1, b_2, ... , b_n\) be real numbers with \(b_1 ≥ b_2 ≥ ... ≥ b_n ≥ 0\) and let \(A_m = \sum_{k=1}^{m}a_k\). Suppose \(|A_m| ≤ B\) for all \(m\). Then \(\left |\sum_{j=1}^{n}a_jb_j \right | \leq B\cdot b_1\).

Prove Lemma \(\PageIndex{2}\).

Prove Theorem \(\PageIndex{1}\).

**Hint**-
Let \(\epsilon > 0\). Since \(\sum_{n=0}^{\infty }a_nr^n\) converges then by the Cauchy Criterion, there exists \(N\) such that if \(m > n > N\) then \(\left |\sum_{k=n+1}^{m} a_k r^k \right | < \frac{\epsilon }{2}\). Let \(0 ≤ x ≤ r\). By Lemma \(\PageIndex{2}\),

\[\left |\sum_{k=n+1}^{m} a_k x^k \right | = \left |\sum_{k=n+1}^{m} a_k r^k \left (\frac{x}{r} \right )^k \right | \leq \frac{\epsilon }{2}\left (\frac{x}{r} \right )^{n+1} \leq \frac{\epsilon }{2} \nonumber\]

Thus for \(0 ≤ x ≤ r\), \(n > N\),

\[\left |\sum_{k=n+1}^{\infty } a_k x^k \right | = \lim_{n \to \infty }\left |\sum_{k=n+1}^{m} a_k x^k \right | \leq \frac{\epsilon }{2} < \epsilon \nonumber\]

Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum a_n(-r)^n\) converges. Then \(\sum a_nx^n\) converges uniformly on \([-r,0]\).

Prove Corollary \(\PageIndex{1}\).

**Hint**-
Consider \(\sum a_n(-x)^n\).

## Contributor

Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)