Abel's Theorem

For a convergent series ∑an→S\sum a_n \to S, the power series ∑anxn\sum a_n x^n approaches SS as x→1−x \to 1^{-} — Abel's theorem. Even the divergent Grandi series points to 12\frac{1}{2} in this limit, but that is the Abel sum, never "the sum" of the series.

∑an→S ⇒ lim⁡x→1−∑anxn=S\sum a_n \to S \ \Rightarrow\ \lim_{x\to 1^{-}}\sum a_n x^n = S
∑n=0∞(−1)nxn=11+x→x→1−12\sum_{n=0}^{\infty}(-1)^n x^n = \frac{1}{1+x} \xrightarrow[x\to 1^{-}]{} \frac{1}{2}

Part of these learning paths

Examples & Exercises

Prove by summation by parts that if the partial sums Sn=a0+⋯+anS_n = a_0 + \cdots + a_n converge to SS, then lim⁡x→1−∑n=0∞anxn=S\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} a_n x^n = S.

∑n=0Nanxn=SNxN+1+(1−x)∑n=0NSnxn\sum_{n=0}^{N} a_n x^n = S_N x^{N+1} + (1-x)\sum_{n=0}^{N} S_n x^n
∑n=0∞anxn−S=(1−x)∑n=0∞(Sn−S)xn\sum_{n=0}^{\infty} a_n x^n - S = (1-x)\sum_{n=0}^{\infty}(S_n - S)x^n
Convergent example at x=0.9x=0.9: S40(0.9)≈0.6420S_{40}(0.9) \approx 0.6420, closed form ln⁡(1.9)≈0.6419\ln(1.9) \approx 0.6419 — the Abel sum points the same way as the ordinary sum ln⁡2\ln 2.
Show proof
  1. With S−1:=0S_{-1} := 0 we have an=Sn−Sn−1a_n = S_n - S_{n-1}. Since ∑n=0N(Sn−Sn−1)xn=∑n=0NSnxn−x∑n=0N−1Snxn\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 ∑n=0Nanxn=SNxN+1+(1−x)∑n=0NSnxn\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).
  2. For fixed ∣x∣<1|x| < 1, letting N→∞N \to \infty kills the boundary term SNxN+1→0S_N x^{N+1} \to 0 (SNS_N is convergent hence bounded, xN+1→0x^{N+1} \to 0). Thus ∑n=0∞anxn=(1−x)∑n=0∞Snxn\sum_{n=0}^{\infty} a_n x^n = (1-x)\sum_{n=0}^{\infty} S_n x^n — the right side converges absolutely since SnS_n is bounded.
  3. Since (1−x)∑n=0∞xn=1(1-x)\sum_{n=0}^{\infty} x^n = 1, we get ∑anxn−S=(1−x)∑n=0∞(Sn−S)xn\sum a_n x^n - S = (1-x)\sum_{n=0}^{\infty}(S_n - S)x^n. Given ε>0\varepsilon > 0, pick N0N_0 with ∣Sn−S∣<ε/2|S_n - S| < \varepsilon/2 for all n>N0n > N_0, and split the sum at N0N_0.
  4. Tail: ∣(1−x)∑n>N0(Sn−S)xn∣≤(1−x)ε2∑n>N0xn=ε2xN0+1≤ε2\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≤x<10 \le x < 1).
  5. Head: ∣(1−x)∑n=0N0(Sn−S)xn∣≤(1−x)∑n=0N0∣Sn−S∣=(1−x)M\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 xn≤1x^n \le 1). MM is a fixed constant, so this tends to 00 as x→1−x \to 1^{-}. For xx close enough to 11 the total is below ε\varepsilon. Hence lim⁡x→1−∑anxn=S\lim_{x \to 1^{-}} \sum a_n x^n = S. ∎

Prove the Abel sum of Grandi's series ∑n=0∞(−1)n\sum_{n=0}^{\infty} (-1)^n is 12\frac{1}{2}. Be explicit that the series itself diverges.

∑n=0N(−x)n=1−(−x)N+11+x→N→∞11+x(∣x∣<1)\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→1−11+x=12\lim_{x\to 1^{-}}\frac{1}{1+x} = \frac{1}{2}
x=0.99x=0.99, N=100N=100: S100(0.99)≈0.6846S_{100}(0.99) \approx 0.6846, closed form 1/1.99≈0.50251/1.99 \approx 0.5025 — raising NN and sending x→1−x \to 1^{-} drives it to 1/21/2.
Show proof
  1. First, honestly: ∑n=0∞(−1)n\sum_{n=0}^{\infty} (-1)^n diverges. The partial sums Sk=1S_k = 1 (kk even), 00 (kk 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.
  2. For ∣x∣<1|x| < 1 the geometric formula gives ∑n=0N(−x)n=1−(−x)N+11+x\sum_{n=0}^{N} (-x)^n = \frac{1-(-x)^{N+1}}{1+x}. E.g. at x=0.9x = 0.9, N=20N = 20: S20(0.9)=1−(−0.9)211.9≈0.5839S_{20}(0.9) = \frac{1-(-0.9)^{21}}{1.9} \approx 0.5839, with exact error ∣(−0.9)21∣1.9≈0.0576\frac{|(-0.9)^{21}|}{1.9} \approx 0.0576.
  3. For ∣x∣<1|x| < 1, (−x)N+1→0(-x)^{N+1} \to 0, so letting N→∞N \to \infty gives ∑n=0∞(−1)nxn=11+x\sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{1+x}. At x=0.9x = 0.9 this is 1/1.9≈0.52631/1.9 \approx 0.5263 — the true target of the S20S_{20} above.
  4. Now let x→1−x \to 1^{-}: 11+x→12\frac{1}{1+x} \to \frac{1}{2}. Hence the Abel sum lim⁡x→1−∑n=0∞(−1)nxn=12\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{2}. It coincides with the Cesàro sum 12\frac{1}{2}, but neither is "the sum" of a divergent series. ∎

Apply Abel's theorem to the alternating harmonic series ∑n=0∞(−1)nn+1=ln⁡2\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \ln 2 to prove lim⁡x→1−∑n=0∞(−1)nxn+1n+1=ln⁡2\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = \ln 2.

∑n=0N(−1)nxn+1n+1=ln⁡(1+x)−RN(x),∣RN(x)∣≤xN+2N+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→1−ln⁡(1+x)=ln⁡2≈0.6931\lim_{x\to 1^{-}}\ln(1+x) = \ln 2 \approx 0.6931
x=0.99x=0.99, N=100N=100: S100(0.99)≈0.6899S_{100}(0.99) \approx 0.6899, closed form ln⁡(1.99)≈0.6881\ln(1.99) \approx 0.6881 — tending to ln⁡2≈0.6931\ln 2 \approx 0.6931 as x→1−x \to 1^{-}.
Show proof
  1. Set an=(−1)nn+1a_n = \frac{(-1)^n}{n+1}. Then ∑n=0∞an\sum_{n=0}^{\infty} a_n converges by the Leibniz test (alternating, 1n+1↘0\frac{1}{n+1} \searrow 0), and its value is ln⁡2≈0.6931\ln 2 \approx 0.6931 (numerical check: at N=100000N = 100000 the partial sum is ≈0.69315218\approx 0.69315218, error ≈5.0×10−6\approx 5.0 \times 10^{-6}).
  2. Abel's theorem (Problem 1) gives lim⁡x→1−∑n=0∞anxn=ln⁡2\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} a_n x^n = \ln 2. But ∑n=0∞(−1)nxn+1n+1=x⋅∑n=0∞anxn\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x \cdot \sum_{n=0}^{\infty} a_n x^n, and x→1x \to 1, so the desired limit is 1⋅ln⁡2=ln⁡21 \cdot \ln 2 = \ln 2.
  3. A direct check agrees: integrating ∑n=0N(−t)n=1−(−t)N+11+t\sum_{n=0}^{N} (-t)^n = \frac{1-(-t)^{N+1}}{1+t} (for ∣t∣<1|t| < 1) from 00 to xx gives ∑n=0N(−1)nxn+1n+1=ln⁡(1+x)−RN(x)\sum_{n=0}^{N} \frac{(-1)^n x^{n+1}}{n+1} = \ln(1+x) - R_N(x) with ∣RN(x)∣≤∫0xtN+1 dt=xN+2N+2→0|R_N(x)| \le \int_0^x t^{N+1}\,dt = \frac{x^{N+2}}{N+2} \to 0 (N→∞N \to \infty). So the power series sums to exactly ln⁡(1+x)\ln(1+x).
  4. Since ln⁡(1+x)→ln⁡2\ln(1+x) \to \ln 2 (x→1−x \to 1^{-}), the limit is ln⁡2≈0.6931\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. ∎

7. Power Series