Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ 1^\circ\; \sum_{n=1}^{\infty} (\sqrt2-\sqrt[3]2)(\sqrt2-\sqrt[5]2)\cdots(\sqrt2-\sqrt[2n+1]2); \qquad 2^\circ\; \sum_{n=1}^{\infty} (\sqrt5-\sqrt[3]5)(\sqrt5-\sqrt[5]5)\cdots(\sqrt5-\sqrt[2n+1]5). $$
Explanation of Method
Both series have the same form: a running product of factors $\sqrt a-a^{1/(2k+1)}$. The ratio $a_{n+1}/a_n$ is just the new factor being multiplied in, and since $a^{1/(2n+3)}\to1$ as $n\to\infty$, the limiting ratio is $\sqrt a-1$ — comparing this to $1$ gives a critical value of $a=4$ that separates convergence from divergence.
- Compute the ratio $a_{n+1}/a_n$ as the newest factor in the product.
- Evaluate the limit, showing it equals $\sqrt a-1$.
- Apply the ratio test separately for $a=2$ and $a=5$.
Step 1 · Compute the ratio
With $a_n=\prod_{k=1}^n\left(a^{1/2}-a^{1/(2k+1)}\right)$:
$$ \frac{a_{n+1}}{a_n} = a^{1/2} - a^{1/(2n+3)}. $$
Step 2 · Evaluate the limit
As $n\to\infty$, $\frac1{2n+3}\to0$, so $a^{1/(2n+3)}\to a^0=1$, giving:
$$ L = \sqrt a - 1. $$
Convergence requires $L<1$, i.e. $\sqrt a<2$, i.e. $a<4$.
Step 3 · Apply the test to each part
Part $1^\circ$ ($a=2$): $L=\sqrt2-1\approx0.414<1$; converges.
Part $2^\circ$ ($a=5$): $L=\sqrt5-1\approx1.236>1$; diverges.
Summary Table
| Part | $a$ | $L=\sqrt a-1$ | Conclusion |
|---|---|---|---|
| $1^\circ$ | $2$ | $\approx0.414$ | Converges |
| $2^\circ$ | $5$ | $\approx1.236$ | Diverges |
Additional Notes
- The critical value $a=4$ (where $\sqrt a=2$) is exact and easy to check: $\sqrt4-1=1$ lands exactly on the ratio test's inconclusive boundary, so any $a$ strictly between $0$ and $4$ converges, and any $a>4$ diverges.
- Each individual factor $a^{1/2}-a^{1/(2k+1)}$ tends to the same limit $\sqrt a-1$ as $k\to\infty$, so the entire product behaves asymptotically like a geometric sequence with ratio $\sqrt a-1$ once $k$ is large — the ratio test result is really just recognizing this eventual geometric-like behavior.
- Since $5>4>2$, the two parts of this problem were deliberately chosen to straddle the critical threshold, illustrating both possible outcomes with concrete numerical values.
Result
Part $1^\circ$ ($a=2$) converges; Part $2^\circ$ ($a=5$) diverges: