Tauber's Theorem

Reduce to $L = 0$: replacing $a_0$ by $a_0 - L$ turns the partial sums into $s_n - L$ (target $0$), the Abel sum into $A(r) - L \to 0$, and leaves $n\,a_n \to 0$ for $n \ge 1$ untouched. Assume henceforth $A(r) \to 0$.

Abel summation by parts for the partial sums $s_n = a_0 + \cdots + a_n$ ($s_{-1} = 0$): for $0 \le r < 1$, $A(r) = (1-r)\sum_{n \ge 0} s_n r^n$.

Justification: since $n\,a_n \to 0$, $(a_n)$ is bounded and $s_N = O(N)$; hence $\sum s_n r^n$ converges absolutely for $|r| < 1$ ($\sum N|r|^N < \infty$) and the boundary term satisfies $s_N r^N \to 0$ (exponential decay beats polynomial growth).

Under these facts, substituting $a_n = s_n - s_{n-1}$ and rearranging $\sum (s_n - s_{n-1}) r^n = (1-r)\sum s_n r^n$ is legitimate.

Given $\varepsilon > 0$, pick $M$ with $|n\,a_n| \le \varepsilon$ for $n \ge M$ (since $n\,a_n \to 0$). For $n, N \ge M$: $|s_N - s_n| = |\sum a_k| \le \varepsilon\sum \frac{1}{k} \le \varepsilon\,|\ln(n/N)|$ (using $\frac{1}{k} \le \int_{k-1}^{k}\frac{dx}{x}$).

Since $(1-r)\sum r^n = 1$, $s_N - A(r) = (1-r)\sum_{n \ge 0}(s_N - s_n)\,r^n$.

Put $r = r_N = 1 - \frac{1}{N}$ and split at $M$.

For $n < M$: $|s_N| \le |s_M| + \varepsilon\ln(N/M)$, so the contribution is $\frac{1}{N}\,M\,(|s_N| + \max_{n<M}|s_n|) \to 0$ as $N \to \infty$.

For $n \ge M$ the contribution is at most $\varepsilon\cdot\frac{1}{N}\sum_{n \ge M}|\ln(n/N)|\left(1-\frac{1}{N}\right)^n$, since $(1 - r_N) = \frac{1}{N}$.

Dominated argument: with $u = n/N$, the singularity of $|\ln u|$ at $u = 0$ is integrable ($\int_0^1 |\ln u|\,du = 1$) on the range $n \le N$, while for $n > N$, $\left(1-\frac{1}{N}\right)^n \le e^{-n/(2N)}$ ($N \ge 2$), so the tail is dominated by the integrable envelope $|\ln u|\,e^{-u/2}$; the finitely many indices $n < M$ are handled separately.

Hence, by dominated convergence for Riemann sums, this expression converges to $\int_0^{\infty} |\ln u|\,e^{-u}\,du$ as $N \to \infty$, so the contribution converges to $\varepsilon\,C_0$ with the finite constant $C_0 = \int_0^{\infty} |\ln u|\,e^{-u}\,du$.

Since $r_N \to 1$ and the Abel sum is $0$, $|A(r_N)| \to 0$.

Hence $\limsup_{N \to \infty} |s_N| \le C_0\,\varepsilon$, and $\varepsilon > 0$ being arbitrary gives $s_N \to 0$ — i.e. $\sum a_n$ converges to $L$.

The heart of the proof: the hypothesis $n\,a_n \to 0$ guarantees $s_n \approx s_N$ wherever the $r_N$-weights concentrate (near $n = N$).

For $|r| < 1$, $\sum_{n=0}^{\infty} (-1)^n r^n = \sum_{n=0}^{\infty} (-r)^n = \frac{1}{1-(-r)} = \frac{1}{1+r}$ (geometric series). As $r \to 1^{-}$, $A(r) = \frac{1}{1+r} \to \frac{1}{2}$ — Grandi's series is Abel summable to $\frac{1}{2}$.

Partial sums: $S_N = 1 - 1 + \cdots \pm 1$ has $N+1$ (odd) terms for even $N$, so $S_N = 1$, and $S_N = 0$ for odd $N$. Two values bouncing forever means no limit, so the series diverges — $\frac{1}{2}$ is never "the sum" of this series.

Since $a_n = (-1)^n$, $n\,a_n = n\,(-1)^n$, i.e. $|n\,a_n| = n \to \infty$.

Tauber's hypothesis $a_n = o(1/n)$ fails, so the theorem promising nothing is no contradiction — and indeed the conclusion fails too (divergence).

Without the hypothesis, Abel summability cannot force convergence.

$|n\,d_n| = \frac{n}{(n+1)\ln(n+2)} = \frac{1}{1 + 1/n}\cdot\frac{1}{\ln(n+2)} \to 1\cdot 0 = 0$ — in one line, $d_n = o(1/n)$.

$|d_n| = \frac{1}{(n+1)\ln(n+2)}$ decreases monotonically to $0$ since the denominator increases in $n$. By the alternating series (Leibniz) test, $\sum d_n$ converges (conditionally) — numerically $S_D \approx 1.13645$ (20000 terms), $|S_{4000} - S_{2000}| \approx 1.8\times 10^{-5}$.

For each fixed $r < 1$, $\sum |d_n|\,r^n$ converges (geometric majorant), so $A_D(r)$ is defined — but that alone does not give $A_D(r) \to L$.

Since $\sum d_n$ was already shown to converge to $S_D$, Abel's continuity theorem gives $A_D(r) \to S_D$ as $r \to 1^{-}$ — i.e. the Abel sum equals the ordinary sum $S_D$.

(This is Abel continuity for an already-convergent series, not Tauber's theorem.)

Numerical check: $A_D(0.999) \approx 1.13666$ differs from $S_{4000} \approx 1.13643$ by only $\approx 2.3\times 10^{-4}$ — exactly the agreement the theorem promises, visible in the chart.