Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{x^n}{n^y}, \qquad x>0,\ y>0. $$
Explanation of Method
The ratio $\frac{a_{n+1}}{a_n}=x\left(\frac n{n+1}\right)^y$ has limit exactly $x$, since the polynomial factor $\left(\frac n{n+1}\right)^y\to1$ regardless of $y$. So the ratio test decides convergence purely by comparing $x$ to $1$ — except at the boundary $x=1$, where the series collapses to an ordinary $p$-series and $y$ finally takes over.
- Compute the ratio $a_{n+1}/a_n$ and its limit, showing it equals $x$.
- Apply the ratio test for $x\neq1$.
- Resolve the boundary case $x=1$ using the $p$-series test on $y$.
Step 1 · Compute the ratio
With $a_n=\dfrac{x^n}{n^y}$:
$$ \frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(n+1)^y}\cdot\frac{n^y}{x^n} = x\left(\frac n{n+1}\right)^y. $$
Since $\frac n{n+1}\to1$, this tends to $L=x\cdot1^y=x$.
Step 2 · Apply the ratio test for $x\neq1$
If $x<1$: $L<1$, converges. If $x>1$: $L>1$, diverges. If $x=1$: $L=1$, inconclusive.
Step 3 · Resolve the boundary case $x=1$
At $x=1$, the series is exactly the $p$-series $\sum\frac1{n^y}$, convergent iff $y>1$.
Summary Table
| Case | Convergence |
|---|---|
| $0<x<1$ | Converges (any $y>0$) |
| $x=1,\ y>1$ | Converges |
| $x=1,\ y\le1$ | Diverges |
| $x>1$ | Diverges (any $y>0$) |
Additional Notes
- The exponent $y$ is entirely irrelevant unless $x=1$ exactly: exponential behavior ($x\ne1$) always dominates polynomial behavior in the denominator, so $y$ can only matter when the exponential factor is neutralized.
- For $x<1$, $\sum x^n/n$ is closely related to the Taylor series $-\ln(1-x)=\sum_{n=1}^\infty x^n/n$, showing this family of series connects directly to the natural logarithm function inside its radius of convergence.
- Problem 59's $\sum2^n/n$ is the special case $x=2>1$, $y=1$ here, confirming its divergence regardless of how large $y$ might otherwise be chosen — e.g. even $\sum2^n/n^{10}$ diverges.
Result
The convergence of the series depends on $x$ and (only at the boundary) on $y$: