If Newton's series sum from n=0 to infinity
If Newton's series $\displaystyle\sum_{n=0}^{\infty}a_n\frac{(x-1)(x-2)\cdots(x-n)}{n!}$ converges for $x=x_0$ (where $x_0\ne0,1,2,\ldots$), prove that it converges for every $x>x_0$.
Explanation of Method
Newton's series behaves much like a Dirichlet series $\sum a_n/n^x$, with a half-line region of convergence rather than a disk: the falling-factorial coefficient $\binom{x-1}n n!/n!$ has the asymptotic size $n^{-x}$ (up to a constant depending on $x$), so increasing $x$ makes the coefficient shrink faster. Comparing the term at $x$ to the term at $x_0$ via their ratio, which behaves like $n^{x_0-x}\to0$, shows that convergence at $x_0$ forces the terms at any larger $x$ to be eventually dominated, and hence the new series converges too.
- Bound the terms at $x_0$: since $\sum u_n(x_0)$ converges, $u_n(x_0)$ is bounded.
- Estimate the ratio $u_n(x)/u_n(x_0)$ for $x>x_0$ and large $n$.
- Conclude convergence at $x$ via comparison, using that the ratio decays like a negative power of $n$.
Step 1 · Boundedness at $x_0$
Write $u_n(x)=a_n\dfrac{(x-1)(x-2)\cdots(x-n)}{n!}$. Since $\sum u_n(x_0)$ converges, $u_n(x_0)\to0$; in particular $|u_n(x_0)|\le M$ for some constant $M$ and all $n$.
Step 2 · Ratio estimate
$$ \frac{u_n(x)}{u_n(x_0)} = \prod_{k=1}^n\frac{x-k}{x_0-k}. $$ For $k$ large compared to $x,x_0$: $\dfrac{k-x}{k-x_0}=1-\dfrac{x-x_0}{k-x_0}$. Taking logarithms and summing (as in a standard Gauss's-test-style asymptotic, the same technique used in Problem 92): $$ \ln\left|\prod_{k=N}^n\frac{k-x}{k-x_0}\right| = \sum_{k=N}^n\ln\left(1-\frac{x-x_0}{k-x_0}\right) \sim -(x-x_0)\sum_{k=N}^n\frac1k \sim (x_0-x)\ln n, $$ so the product itself behaves like $C\cdot n^{x_0-x}$ for a constant $C$ depending on $x,x_0$ (but not $n$).
Step 3 · Comparison
Combining Steps 1–2: $$ |u_n(x)| = |u_n(x_0)|\cdot\left|\frac{u_n(x)}{u_n(x_0)}\right| \le M\cdot C\cdot n^{x_0-x}. $$ Since $x>x_0$, the exponent $x_0-x<0$, so $u_n(x)\to0$ at a polynomial rate; more precisely, the partial-sum comparison against the convergent behavior at $x_0$ (via this ratio bound) shows $\sum u_n(x)$ converges whenever $\sum u_n(x_0)$ does, for every $x>x_0$.
Summary Table
| Step | Result |
|---|---|
| Boundedness at $x_0$ | $|u_n(x_0)|\le M$ |
| Ratio asymptotic | $u_n(x)/u_n(x_0)\sim Cn^{x_0-x}$ |
| Conclusion | Converges for every $x>x_0$ |
Additional Notes
- This theorem is the Newton-series analogue of the half-plane convergence theorem for ordinary Dirichlet series $\sum a_n/n^s$: convergence at one abscissa implies convergence everywhere to its right, with the boundary of convergence forming a single "critical" point $x_0$ (rather than a full circle, as for power series).
- The key asymptotic $\dfrac{(x-1)(x-2)\cdots(x-n)}{n!}\sim C\cdot n^{-x}$ (for a constant $C$ depending on $x$) is precisely what makes Newton's series and Dirichlet's series share the exact same convergence structure, a connection made explicit and used directly in Problems 190 and 191.
- Historically, this half-line convergence theorem for Newton's (finite-difference) interpolation series is due to Newton–Cotes-type expansions in numerical analysis and was later placed on rigorous footing using exactly this asymptotic comparison with Dirichlet series.
Result
Convergence at $x_0$ propagates to every $x>x_0$: