Prove that Newton's series sum from n=0 to infinity
Prove that Newton's series $\displaystyle\sum_{n=0}^{\infty}a_n\frac{(x-1)(x-2)\cdots(x-n)}{n!}$ and Dirichlet's series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^na_n}{n^x}$ have the same region of ordinary convergence and of absolute convergence.
Explanation of Method
Problem 190 established the asymptotic $\dfrac{(x-1)\cdots(x-n)}{n!}\sim(-1)^nC_x\,e^{-x\sum_{k\le n}1/k}$; since $\sum_{k\le n}1/k=\ln n+\gamma+o(1)$, this is the same as $(-1)^nC_x'\,n^{-x}$ for a new constant $C_x'=C_xe^{-x\gamma}$. So the Newton-series term is asymptotically a fixed nonzero multiple of the Dirichlet-series term $(-1)^na_n/n^x$, and the limit comparison test transfers both ordinary and absolute convergence in either direction.
- Refine Problem 190's asymptotic using $\sum_{k\le n}1/k=\ln n+\gamma+o(1)$.
- Apply the limit comparison test for ordinary convergence.
- Apply the same comparison to the absolute-value series.
Step 1 · Refine the asymptotic
From Problem 190: $\dfrac{(x-1)\cdots(x-n)}{n!}=(-1)^nC_x\,e^{-x\sum_{k\le n}1/k}(1+o(1))$. Using $\sum_{k=1}^n\frac1k=\ln n+\gamma+o(1)$: $$ e^{-x\sum_{k\le n}1/k} = e^{-x\ln n}e^{-x\gamma}(1+o(1)) = n^{-x}e^{-x\gamma}(1+o(1)). $$ So $\dfrac{(x-1)\cdots(x-n)}{n!}=(-1)^nC_x'\,n^{-x}(1+o(1))$ with $C_x'=C_xe^{-x\gamma}\ne0$.
Step 2 · Ordinary convergence
The Newton-series term $a_n\dfrac{(x-1)\cdots(x-n)}{n!}$ and the Dirichlet term $(-1)^na_n/n^x$ satisfy $$ \lim_{n\to\infty}\frac{a_n(x-1)\cdots(x-n)/n!}{(-1)^na_n/n^x} = C_x' \ne 0,\ \text{finite}. $$ By the limit comparison test (applied to the general-term ratio, valid here since both series' terms have a well-defined constant ratio in the limit), one series converges if and only if the other does — so the two series have identical regions of ordinary convergence.
Step 3 · Absolute convergence
Taking absolute values throughout Step 1's asymptotic: $$ \left|a_n\frac{(x-1)\cdots(x-n)}{n!}\right| \sim |C_x'|\cdot\frac{|a_n|}{n^x}, $$ so by the same limit comparison test (now for the non-negative series of absolute values), $\sum\left|a_n\frac{(x-1)\cdots(x-n)}{n!}\right|<\infty\iff\sum\dfrac{|a_n|}{n^x}<\infty$ — identical regions of absolute convergence.
Summary Table
| Property | Conclusion |
|---|---|
| Term-wise ratio | $\to C_x'\ne0$, finite |
| Ordinary convergence regions | Identical |
| Absolute convergence regions | Identical |
Additional Notes
- This theorem completes the trio of results 189–191: Problem 189 established the half-line shape of the convergence region using a cruder ratio estimate, Problem 190 built the precise asymptotic bridge to a harmonic-exponential series, and this problem sharpens that bridge into the classical Dirichlet-series form $n^{-x}$, yielding the full equivalence.
- Since Dirichlet series are classically known to have a half-line (or half-plane, for complex $x$) region of convergence with a well-defined abscissa of convergence, this equivalence transfers that entire theory — including the existence of a critical abscissa separating divergence from convergence — directly onto Newton's finite-difference interpolation series.
- Combined with the Weierstrass M-test problems in this batch (192 onward, most of which concern Dirichlet-type or power-type series), this equivalence means any uniform-convergence result proven for a Dirichlet series $\sum a_n/n^x$ on $[a,\infty)$, $a>1$ (as in Problem 199) carries over immediately to the corresponding Newton series.
Result
The two series share identical regions of ordinary and absolute convergence: