Using the Cauchy criterion, test the uniform convergence
Using the Cauchy criterion, test the uniform convergence of the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{nx}{(1+x)(1+2x)\cdots(1+nx)}$ on $[1,\infty)$.
Explanation of Method
The term telescopes exactly: it equals the difference of reciprocal partial products at consecutive stages, so the partial sum collapses to $1-1/\prod_{k=1}^N(1+kx)$. Since $x\ge1$ makes each factor $(1+kx)\ge1+k$, the product is bounded below by $(N+1)!$, giving an extremely fast (factorial) uniform bound on the remainder.
- Show the term telescopes and find the closed-form partial sum.
- Bound the product $\prod(1+kx)$ below using $x\ge1$.
- Confirm the Cauchy criterion with an explicit factorial bound.
Step 1 · Telescoping structure
With $P_n(x)=\prod_{k=1}^n(1+kx)$ (and $P_0=1$): $$ \frac1{P_{n-1}(x)}-\frac1{P_n(x)} = \frac{(1+nx)-1}{P_n(x)} = \frac{nx}{P_n(x)}, $$ exactly the $n$-th term. So $S_N(x)=\sum_{n=1}^N\dfrac{nx}{P_n(x)}=1-\dfrac1{P_N(x)}$, and as $N\to\infty$ (since $P_N(x)\to\infty$ for $x>0$), $S(x)\equiv1$.
Step 2 · Bound the product for $x\ge1$
$P_N(x)=\prod_{k=1}^N(1+kx)\ge\prod_{k=1}^N(1+k)=(N+1)!$ for $x\ge1$.
Step 3 · Cauchy criterion
For $n>m$ and $x\ge1$: $$ \left|\sum_{k=m+1}^n\frac{kx}{P_k(x)}\right| = \frac1{P_m(x)}-\frac1{P_n(x)} \le \frac1{P_m(x)} \le \frac1{(m+1)!}. $$ Given $\epsilon>0$, choose $N$ with $1/(N+1)!<\epsilon$; the Cauchy criterion then holds uniformly for all $n>m\ge N$ and all $x\ge1$.
Summary Table
| Quantity | Value |
|---|---|
| Partial sum $S_N(x)$ | $1-1/P_N(x)$ |
| Sum $S(x)$ | $1$ (all $x\ge1$) |
| Remainder bound | $\le1/(m+1)!\to0$ |
Additional Notes
- The factorial rate of decay here is by far the fastest uniform-convergence rate in this batch, faster even than the exponential rate of Problem 176 on $[0,a]$ — a direct consequence of the multiplicative (rather than additive) accumulation in the denominator.
- This telescoping technique — recognizing a term as a difference of reciprocal partial products — is the multiplicative analogue of the additive telescoping used throughout Problems 141, 149, 153–157, and 186; both rely on finding an antiderivative-like closed form for the partial sum before estimating the remainder.
- The restriction $x\ge1$ (rather than $x>0$) is what makes the clean factorial bound $(N+1)!$ available; for $x$ near $0$, the product $P_N(x)$ grows much more slowly, and the convergence, while still uniform on any $[a,\infty)$ with $a>0$ by the same argument, would only be geometric-like rather than factorial in rate.
Result
The series converges uniformly to $1$ on $[1,\infty)$, with factorial rate: