Tauber's Theorem

Grandi's series has Abel sum 12\frac{1}{2} yet diverges — because ∣n an∣=n|n\,a_n| = n breaks the hypothesis an=o(1/n)a_n = o(1/n). The series dn=(−1)n(n+1)ln⁡(n+2)d_n = \frac{(-1)^n}{(n+1)\ln(n+2)} satisfies it, and its Abel-sum curve reaches the ordinary sum — Tauber's theorem.

A(r)=∑anrn→r→1−L, an=o(1/n) ⇒ ∑an=LA(r) = \sum a_n r^n \xrightarrow[r\to 1^{-}]{} L,\ a_n = o(1/n)\ \Rightarrow\ \sum a_n = L
∑n=0∞(−1)nrn=11+r→r→1−12\sum_{n=0}^{\infty}(-1)^n r^n = \frac{1}{1+r} \xrightarrow[r\to 1^{-}]{} \frac{1}{2}

Part of these learning paths

Examples & Exercises

Prove Tauber's theorem completely: if ∑an\sum a_n is Abel summable to LL and an=o(1/n)a_n = o(1/n), then ∑an\sum a_n converges to LL.

A(r)=∑n≥0anrn→r→1−L,n an→0 ⇒ sN=∑n≤Nan→LA(r) = \sum_{n \ge 0} a_n r^n \xrightarrow[r \to 1^{-}]{} L, \quad n\,a_n \to 0 \ \Rightarrow\ s_N = \sum_{n \le N} a_n \to L
At r=0.9r = 0.9, Grandi's AG=1/1.9≈0.5263A_G = 1/1.9 \approx 0.5263 heads toward 12\frac{1}{2}, but ∣n an∣=n|n\,a_n| = n breaks the hypothesis of the proof below — so the conclusion breaks too.
Show proof
  1. Reduce to L=0L = 0: replacing a0a_0 by a0−La_0 - L turns the partial sums into sn−Ls_n - L (target 00), the Abel sum into A(r)−L→0A(r) - L \to 0, and leaves n an→0n\,a_n \to 0 for n≥1n \ge 1 untouched. Assume henceforth A(r)→0A(r) \to 0.
  2. Abel summation by parts for the partial sums sn=a0+⋯+ans_n = a_0 + \cdots + a_n (s−1=0s_{-1} = 0): for 0≤r<10 \le r < 1, A(r)=(1−r)∑n≥0snrnA(r) = (1-r)\sum_{n \ge 0} s_n r^n. (Substitute an=sn−sn−1a_n = s_n - s_{n-1} and ∑(sn−sn−1)rn=(1−r)∑snrn\sum (s_n - s_{n-1}) r^n = (1-r)\sum s_n r^n — the rearrangement is legitimate for ∣r∣<1|r| < 1.)
  3. Given ε>0\varepsilon > 0, pick MM with ∣n an∣≤ε|n\,a_n| \le \varepsilon for n≥Mn \ge M (since n an→0n\,a_n \to 0). For n,N≥Mn, N \ge M: ∣sN−sn∣=∣∑ak∣≤ε∑1k≤ε ∣ln⁡(n/N)∣|s_N - s_n| = |\sum a_k| \le \varepsilon\sum \frac{1}{k} \le \varepsilon\,|\ln(n/N)| (using 1k≤∫k−1kdxx\frac{1}{k} \le \int_{k-1}^{k}\frac{dx}{x}).
  4. Since (1−r)∑rn=1(1-r)\sum r^n = 1, sN−A(r)=(1−r)∑n≥0(sN−sn) rns_N - A(r) = (1-r)\sum_{n \ge 0}(s_N - s_n)\,r^n. Put r=rN=1−1Nr = r_N = 1 - \frac{1}{N} and split at MM. For n<Mn < M: ∣sN∣≤∣sM∣+εln⁡(N/M)|s_N| \le |s_M| + \varepsilon\ln(N/M), so the contribution is 1N M (∣sN∣+max⁡n<M∣sn∣)→0\frac{1}{N}\,M\,(|s_N| + \max_{n<M}|s_n|) \to 0 as N→∞N \to \infty.
  5. For n≥Mn \ge M the contribution is at most ε(1−1N)∑n≥M∣ln⁡(n/N)∣(1−1N)n\varepsilon\left(1-\frac{1}{N}\right)\sum_{n \ge M}|\ln(n/N)|\left(1-\frac{1}{N}\right)^n. Substituting u=n/Nu = n/N turns this into a Riemann sum for ∫0∞∣ln⁡u∣ e−u du\int_0^{\infty} |\ln u|\,e^{-u}\,du (the integrand is integrable), so it converges to ε C0\varepsilon\,C_0 (a finite constant) as N→∞N \to \infty.
    (1−1N)∑n≥M∣ln⁡nN∣(1−1N)n⟶∫0∞∣ln⁡u∣ e−u du<∞\left(1-\frac{1}{N}\right)\sum_{n \ge M} \left\lvert\ln\frac{n}{N}\right\rvert \left(1-\frac{1}{N}\right)^n \longrightarrow \int_0^{\infty} \lvert\ln u\rvert\,e^{-u}\,du < \infty
  6. Since rN→1r_N \to 1 and the Abel sum is 00, ∣A(rN)∣→0|A(r_N)| \to 0. Hence lim sup⁡N→∞∣sN∣≤C0 ε\limsup_{N \to \infty} |s_N| \le C_0\,\varepsilon, and ε>0\varepsilon > 0 being arbitrary gives sN→0s_N \to 0 — i.e. ∑an\sum a_n converges to LL. The heart of the proof: the hypothesis n an→0n\,a_n \to 0 guarantees sn≈sNs_n \approx s_N wherever the rNr_N-weights concentrate (near n=Nn = N).

For Grandi's series 1−1+1−⋯1 - 1 + 1 - \cdots, compute A(r)=11+r→12A(r) = \frac{1}{1+r} \to \frac{1}{2}, show the partial sums diverge by oscillation, and show n an↛0n\,a_n \not\to 0 — so both the hypothesis and the conclusion of Tauber's theorem fail.

A(r)=∑n=0∞(−1)nrn=11+r→r→1−12A(r) = \sum_{n=0}^{\infty} (-1)^n r^n = \frac{1}{1+r} \xrightarrow[r \to 1^{-}]{} \frac{1}{2}
SN={1N even0N odd,∣N aN∣=N↛0S_N = \begin{cases} 1 & N\ \mathrm{even} \\ 0 & N\ \mathrm{odd} \end{cases}, \qquad \lvert N\,a_N\rvert = N \not\to 0
r=0.9r = 0.9: AG=1/1.9≈0.5263A_G = 1/1.9 \approx 0.5263 heads toward 12\frac{1}{2}, yet the partial sums still bounce 1,0,1,0…1, 0, 1, 0\ldots.
Show proof
  1. For ∣r∣<1|r| < 1, ∑n=0∞(−1)nrn=∑n=0∞(−r)n=11−(−r)=11+r\sum_{n=0}^{\infty} (-1)^n r^n = \sum_{n=0}^{\infty} (-r)^n = \frac{1}{1-(-r)} = \frac{1}{1+r} (geometric series). As r→1−r \to 1^{-}, A(r)=11+r→12A(r) = \frac{1}{1+r} \to \frac{1}{2} — Grandi's series is Abel summable to 12\frac{1}{2}.
  2. Partial sums: SN=1−1+⋯±1S_N = 1 - 1 + \cdots \pm 1 has N+1N+1 (odd) terms for even NN, so SN=1S_N = 1, and SN=0S_N = 0 for odd NN. Two values bouncing forever means no limit, so the series diverges — 12\frac{1}{2} is never "the sum" of this series.
  3. Since an=(−1)na_n = (-1)^n, n an=n (−1)nn\,a_n = n\,(-1)^n, i.e. ∣n an∣=n→∞|n\,a_n| = n \to \infty. Tauber's hypothesis an=o(1/n)a_n = o(1/n) fails, so the theorem promising nothing is no contradiction — and indeed the conclusion fails too (divergence). Without the hypothesis, Abel summability cannot force convergence.

For dn=(−1)n(n+1)ln⁡(n+2)d_n = \frac{(-1)^n}{(n+1)\ln(n+2)}, prove n dn→0n\,d_n \to 0 (one line) and prove ∑dn\sum d_n converges (alternating series test). Then apply Tauber's theorem to conclude its Abel sum equals its ordinary sum.

∣n dn∣=n(n+1)ln⁡(n+2)→0\lvert n\,d_n\rvert = \frac{n}{(n+1)\ln(n+2)} \to 0
SD≈1.13645,AD(0.999)≈1.13666S_D \approx 1.13645, \qquad A_D(0.999) \approx 1.13666
∣AD(0.999)−SD∣≈2.3×10−4\lvert A_D(0.999) - S_D \rvert \approx 2.3\times 10^{-4}
r=0.9r = 0.9: AD(0.9)≈1.1583A_D(0.9) \approx 1.1583 heads toward the ordinary sum SD≈1.13645S_D \approx 1.13645, and at r=0.999r = 0.999 it is ≈1.13666\approx 1.13666 — nearly exact. The hypothesis holds, so the Abel sum agrees with the true sum.
Show proof
  1. ∣n dn∣=n(n+1)ln⁡(n+2)=11+1/n⋅1ln⁡(n+2)→1⋅0=0|n\,d_n| = \frac{n}{(n+1)\ln(n+2)} = \frac{1}{1 + 1/n}\cdot\frac{1}{\ln(n+2)} \to 1\cdot 0 = 0 — in one line, dn=o(1/n)d_n = o(1/n).
  2. ∣dn∣=1(n+1)ln⁡(n+2)|d_n| = \frac{1}{(n+1)\ln(n+2)} decreases monotonically to 00 since the denominator increases in nn. By the alternating series (Leibniz) test, ∑dn\sum d_n converges (conditionally) — numerically SD≈1.13645S_D \approx 1.13645 (20000 terms), ∣S4000−S2000∣≈1.8×10−5|S_{4000} - S_{2000}| \approx 1.8\times 10^{-5}.
  3. Abel summability of ∑dn\sum d_n is immediate (∑∣dn∣ rn\sum |d_n|\,r^n converges by comparison), and by Tauber's theorem (first problem) its Abel sum equals the ordinary sum SDS_D. Numerical check: AD(0.999)≈1.13666A_D(0.999) \approx 1.13666 differs from S4000≈1.13643S_{4000} \approx 1.13643 by only ≈2.3×10−4\approx 2.3\times 10^{-4} — exactly the agreement the theorem promises, visible in the chart.

7. Power Series