Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}, \qquad a>0. $$
Explanation of Method
Writing $P_n=\prod_{k=1}^n(1+a^k)$, the ratio $a_{n+1}/a_n$ simplifies to $\frac a{1+a^{n+1}}$, since $P_{n+1}=P_n(1+a^{n+1})$. Checking every regime of $a$ — small, equal to $1$, or large — shows the limit is always strictly below $1$, so the growing product in the denominator guarantees convergence no matter how $a$ is chosen.
- Express the denominator as a product $P_n$ and compute the ratio $a_{n+1}/a_n$.
- Evaluate the limit in each regime: $0<a<1$, $a=1$, $a>1$.
- Apply the ratio test, confirming $L<1$ universally.
Step 1 · Compute the ratio
Let $P_n=\prod_{k=1}^n(1+a^k)$, so $a_n=\dfrac{a^n}{P_n}$. Since $P_{n+1}=P_n(1+a^{n+1})$:
$$ \frac{a_{n+1}}{a_n} = a\cdot\frac{P_n}{P_{n+1}} = \frac{a}{1+a^{n+1}}. $$
Step 2 · Evaluate the limit in each regime
$0<a<1$: $a^{n+1}\to0$, so $L=\dfrac a{1+0}=a<1$.
$a=1$: $L=\dfrac1{1+1}=\dfrac12<1$.
$a>1$: $a^{n+1}\to\infty$, so $L=\lim\dfrac a{1+a^{n+1}} = \lim a\cdot a^{-(n+1)} = 0<1$.
Step 3 · Apply the ratio test
In every case, $L<1$, so by D'Alembert's ratio test $\sum_{n=1}^{\infty}\dfrac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}$ converges for all $a>0$.
Summary Table
| Regime | $L=\lim a_{n+1}/a_n$ |
|---|---|
| $0<a<1$ | $a$ |
| $a=1$ | $\dfrac12$ |
| $a>1$ | $0$ |
Additional Notes
- At $a=1$, every factor $(1+1^k)=2$, so $P_n=2^n$ exactly, and the series reduces to the geometric series $\sum\frac1{2^n}=1$ — a clean check that the general machinery correctly recovers a familiar result in this special case.
- For $a>1$, the product $P_n$ grows super-exponentially — roughly like $2^{n(n+1)/2}$ for large $a$ — so the terms $a_n$ decay far faster than any fixed geometric rate, explaining why $L\to0$ rather than merely staying below $1$.
- The universal convergence here (for all $a>0$) parallels Problem 62's $\sum n^p/n!$: in both cases a product-type denominator that grows without any fixed upper bound on its rate is powerful enough to force convergence regardless of the competing growth in the numerator.
Result
The series converges for every $a>0$: