Abel's Theorem

If the series converges, the power series heads to the same value as x nears 1, as in the formula below. Even the divergent Grandi series points to one half in this limit. But that value is an Abel sum, a separate convention, not the sum.

For example, imagine a 🧑 friend shouting 1 and 0 in turns. If each shout is a bit quieter than the last, the running average settles near one half.

Examples & Exercises

Prove by summation by parts that if the partial sums $S_n = a_0 + \cdots + a_n$ converge to $S$ , then $\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} a_n x^n = S$ .

\sum_{n=0}^{N} a_n x^n = S_N x^{N+1} + (1-x)\sum_{n=0}^{N} S_n x^n

\sum_{n=0}^{\infty} a_n x^n - S = (1-x)\sum_{n=0}^{\infty}(S_n - S)x^n

Convergent example at

x=0.9

S_{40}(0.9) \approx 0.6420

, closed form

\ln(1.9) \approx 0.6419

— the Abel sum points the same way as the ordinary sum

\ln 2

Show proof

With $S_{-1} := 0$ we have $a_n = S_n - S_{n-1}$ . Since $\sum_{n=0}^{N} (S_n - S_{n-1})x^n = \sum_{n=0}^{N} S_n x^n - x\sum_{n=0}^{N-1} S_n x^n$ , rearranging gives $\sum_{n=0}^{N} a_n x^n = S_N x^{N+1} + (1-x)\sum_{n=0}^{N} S_n x^n$ — Abel's transformation (summation by parts).
For fixed $|x| < 1$ , letting $N \to \infty$ kills the boundary term $S_N x^{N+1} \to 0$ ( $S_N$ is convergent hence bounded, $x^{N+1} \to 0$ ). Thus $\sum_{n=0}^{\infty} a_n x^n = (1-x)\sum_{n=0}^{\infty} S_n x^n$ — the right side converges absolutely since $S_n$ is bounded.
Since $(1-x)\sum_{n=0}^{\infty} x^n = 1$ , we get $\sum a_n x^n - S = (1-x)\sum_{n=0}^{\infty}(S_n - S)x^n$ . Given $\varepsilon > 0$ , pick $N_0$ with $|S_n - S| < \varepsilon/2$ for all $n > N_0$ , and split the sum at $N_0$ .
Tail: $\left|(1-x)\sum_{n>N_0}(S_n - S)x^n\right| \le (1-x)\frac{\varepsilon}{2}\sum_{n>N_0}x^n = \frac{\varepsilon}{2}x^{N_0+1} \le \frac{\varepsilon}{2}$ (for $0 \le x < 1$ ).
Head: $\left|(1-x)\sum_{n=0}^{N_0}(S_n - S)x^n\right| \le (1-x)\sum_{n=0}^{N_0}|S_n - S| = (1-x)M$ (using $x^n \le 1$ ).
$M$ is a fixed constant, so this tends to $0$ as $x \to 1^{-}$ .
For $x$ close enough to $1$ the total is below $\varepsilon$ .
Hence $\lim_{x \to 1^{-}} \sum a_n x^n = S$ . ∎

Prove the Abel sum of Grandi's series $\sum_{n=0}^{\infty} (-1)^n$ is $\frac{1}{2}$ .

Be explicit that the series itself diverges.

\sum_{n=0}^{N}(-x)^n = \frac{1-(-x)^{N+1}}{1+x} \xrightarrow[N\to\infty]{} \frac{1}{1+x} \quad (|x|<1)

\lim_{x\to 1^{-}}\frac{1}{1+x} = \frac{1}{2}

x=0.99

N=100

S_{100}(0.99) \approx 0.6846

, closed form

1/1.99 \approx 0.5025

— raising

N

and sending

x \to 1^{-}

drives it to

1/2

Show proof

First, honestly: $\sum_{n=0}^{\infty} (-1)^n$ diverges.
The partial sums $S_k = 1$ ( $k$ even), $0$ ( $k$ odd) bounce forever with no limit (see the Grandi scene).
What we compute now is not "the sum" of the series, but its Abel sum.
For $|x| < 1$ the geometric formula gives $\sum_{n=0}^{N} (-x)^n = \frac{1-(-x)^{N+1}}{1+x}$ . E.g. at $x = 0.9$ , $N = 20$ : $S_{20}(0.9) = \frac{1-(-0.9)^{21}}{1.9} \approx 0.5839$ , with exact error $\frac{|(-0.9)^{21}|}{1.9} \approx 0.0576$ .
For $|x| < 1$ , $(-x)^{N+1} \to 0$ , so letting $N \to \infty$ gives $\sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{1+x}$ . At $x = 0.9$ this is $1/1.9 \approx 0.5263$ — the true target of the $S_{20}$ above.
Now let $x \to 1^{-}$ : $\frac{1}{1+x} \to \frac{1}{2}$ .
Hence the Abel sum $\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{2}$ .
It coincides with the Cesàro sum $\frac{1}{2}$ , but neither is "the sum" of a divergent series. ∎

Apply Abel's theorem to the alternating harmonic series $\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \ln 2$ to prove $\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = \ln 2$ .

\sum_{n=0}^{N}\frac{(-1)^n x^{n+1}}{n+1} = \ln(1+x) - R_N(x), \qquad |R_N(x)| \le \frac{x^{N+2}}{N+2}

\lim_{x\to 1^{-}}\ln(1+x) = \ln 2 \approx 0.6931

x=0.99

N=100

S_{100}(0.99) \approx 0.6899

, closed form

\ln(1.99) \approx 0.6881

— tending to

\ln 2 \approx 0.6931

x \to 1^{-}

Show proof

Set $a_n = \frac{(-1)^n}{n+1}$ . Then $\sum_{n=0}^{\infty} a_n$ converges by the Leibniz test (alternating, $\frac{1}{n+1} \searrow 0$ ), and its value is $\ln 2 \approx 0.6931$ (numerical check: at $N = 100000$ the partial sum is $\approx 0.69315218$ , error $\approx 5.0 \times 10^{-6}$ ).
Abel's theorem (Problem 1) gives $\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} a_n x^n = \ln 2$ . But $\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x \cdot \sum_{n=0}^{\infty} a_n x^n$ , and $x \to 1$ , so the desired limit is $1 \cdot \ln 2 = \ln 2$ .
A direct check agrees: restrict to $0 \le x < 1$ .
Then $t$ on the integration interval $[0, x]$ satisfies $0 \le t \le x < 1$ and $1 + t \ge 1$ .
Integrating $\sum_{n=0}^{N} (-t)^n = \frac{1-(-t)^{N+1}}{1+t}$ (for $|t| < 1$ ) from $0$ to $x$ gives $\sum_{n=0}^{N} \frac{(-1)^n x^{n+1}}{n+1} = \ln(1+x) - R_N(x)$ with $|R_N(x)| \le \int_0^x \frac{t^{N+1}}{1+t}\,dt \le \int_0^x t^{N+1}\,dt = \frac{x^{N+2}}{N+2} \to 0$ ( $N \to \infty$ ), using $1 + t \ge 1$ .
So for $0 \le x < 1$ the power series sums to exactly $\ln(1+x)$ .
The case $x = 1$ then follows by Abel's theorem (continuity), not by this integration step.
Since $\ln(1+x) \to \ln 2$ ( $x \to 1^{-}$ ), the limit is $\ln 2 \approx 0.6931$ — matching Abel's theorem's prediction. This shows the Abel sum never contradicts the ordinary sum for a convergent series. ∎

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Examples & Exercises