Prove that if the series sum from n=1
Prove that if the series $\displaystyle\sum_{n=1}^{\infty}(-1)^na_ne^{-\tau\sum_{k=1}^n\frac1k}$ converges, then the series $\displaystyle\sum_{n=0}^{\infty}a_n\frac{(x-1)(x-2)\cdots(x-n)}{n!}$ converges (with $\tau=x$).
Explanation of Method
Rewriting the falling factorial as $(-1)^n$ times a rising product of positive factors $(k-x)$, and using the classical expansion $\sum_{k=1}^n1/k=\ln n+\gamma+o(1)$, shows that $\dfrac{\prod_{k=1}^n(k-x)}{n!}$ is asymptotically equal (up to a bounded factor) to $e^{-x\sum_{k=1}^n1/k}$. This makes the Newton-series term and the given series' term asymptotically proportional, so convergence of one controls convergence of the other by direct comparison.
- Rewrite the falling factorial via $(-1)^n$ and a positive product.
- Expand $\ln\prod_{k=1}^n(1-x/k)$ using the harmonic sum asymptotic $\sum1/k\sim\ln n+\gamma$.
- Compare the two series' terms asymptotically and conclude.
Step 1 · Rewrite via the harmonic-sum exponential
$(x-1)(x-2)\cdots(x-n)=(-1)^n(1-x)(2-x)\cdots(n-x)=(-1)^nn!\displaystyle\prod_{k=1}^n\left(1-\frac xk\right)$, so $$ a_n\frac{(x-1)\cdots(x-n)}{n!} = (-1)^na_n\prod_{k=1}^n\left(1-\frac xk\right). $$
Step 2 · Asymptotic via the harmonic sum
Using $\ln(1-t)=-t+O(t^2)$ for small $t=x/k$: $$ \ln\prod_{k=1}^n\left(1-\frac xk\right) = \sum_{k=1}^n\ln\left(1-\frac xk\right) = -x\sum_{k=1}^n\frac1k + O(1), $$ so $\displaystyle\prod_{k=1}^n\left(1-\frac xk\right) = C_x\cdot e^{-x\sum_{k=1}^n1/k}\cdot(1+o(1))$ for a bounded, eventually-constant factor $C_x$ depending only on $x$.
Step 3 · Comparison
Combining Steps 1–2: $$ a_n\frac{(x-1)\cdots(x-n)}{n!} \sim C_x\cdot(-1)^na_ne^{-x\sum_{k=1}^n1/k}, $$ so the Newton-series term is asymptotically a constant multiple of the given series' term (with $\tau=x$). Since the given series $\sum(-1)^na_ne^{-\tau\sum1/k}$ converges by hypothesis, and the ratio of terms tends to the finite nonzero constant $C_x$, the Newton series converges too.
Summary Table
| Quantity | Asymptotic |
|---|---|
| $\prod_{k=1}^n(1-x/k)$ | $\sim C_x\,e^{-x\sum_{k\le n}1/k}$ |
| Newton-series term | $\sim C_x\cdot$(given series' term) |
Additional Notes
- This problem makes precise, via an explicit asymptotic comparison, the intuitive link between Newton's series and Dirichlet-type series that underlies Problem 189's half-line convergence theorem — the exponential-harmonic form here is essentially $e^{-x\ln n}=n^{-x}$ in disguise, since $\sum_{k\le n}1/k\sim\ln n$.
- This result is the direct stepping stone to Problem 191, which upgrades this one-directional comparison into a full equivalence between the convergence regions of Newton's series and the classical Dirichlet series $\sum(-1)^na_n/n^x$.
- The appearance of the Euler–Mascheroni constant $\gamma$ (implicitly, inside the $O(1)$ term in Step 2) reflects that $\sum_{k\le n}1/k=\ln n+\gamma+o(1)$ is the precise refinement of the harmonic sum's leading asymptotic, though only the leading $\ln n$ behavior is needed for the comparison argument itself.
Result
Convergence of the harmonic-exponential series (at $\tau=x$) implies convergence of Newton's series: