Investigate the convergence of the series (parametrized
Investigate the convergence of the series (parametrized by real $a$ and $b\ge0$): $$ \sum_{n=1}^{\infty} \frac{a^n}{b^n+n}. $$
Explanation of Method
The denominator's dominant term depends on whether $b>1$, $b=1$, or $b<1$, since $b^n$ and $n$ compete differently in each regime. Splitting into these three cases and applying the root/ratio test (or direct comparison) to each resolves absolute convergence, and the single delicate boundary is $a=-1$ with $0\le b\le1$, where the denominator no longer grows geometrically and Leibniz's test must be invoked.
- Compare $b^n+n$ to $\max(b,1)^n$ in each regime of $b$.
- Apply the root test to the absolute series $\sum|a|^n/(b^n+n)$.
- Resolve the boundary case $|a|=\max(b,1)$, especially $a=-1$, $0\le b\le1$.
Step 1 · Denominator asymptotics
If $b>1$: $b^n+n\sim b^n$. If $b=1$: $b^n+n=1+n\sim n$. If $0\le b<1$: $b^n+n\sim n$ (since $b^n\to0$). So for $b\le1$, $b^n+n\sim n$; for $b>1$, $b^n+n\sim b^n$.
Step 2 · Root test on $\sum\left|\dfrac{a^n}{b^n+n}\right|$
For $b>1$: $\sqrt[n]{|a|^n/(b^n+n)}\to|a|/b$, so absolute convergence iff $|a|<b$. For $b\le1$: $\sqrt[n]{|a|^n/n}\to|a|$, so absolute convergence iff $|a|<1$. Combining: absolute convergence iff $|a|<\max(b,1)$.
Step 3 · Boundary cases
If $|a|=\max(b,1)>1$ (i.e. $b>1$, $a=\pm b$): terms $\sim(\pm1)^n/b^n\cdot b^n=(\pm1)^n\not\to0$... more precisely $a^n/(b^n+n)\to\pm1\ne0$ (dividing numerator/denominator by $b^n$), so the series diverges by the divergence test. If $|a|=\max(b,1)=1$ (i.e. $0\le b\le1$, $a=\pm1$): for $a=1$, $\sum1/(b^n+n)\sim\sum1/n$ diverges; for $a=-1$, $(-1)^n/(b^n+n)$ has terms $\to0$ monotonically decreasing in absolute value (denominator increasing), so by Leibniz's test the series converges — conditionally, since $\sum1/(b^n+n)$ diverges.
Summary Table
| Condition | Conclusion |
|---|---|
| $|a|<\max(b,1)$ | Absolutely convergent |
| $a=-1$, $0\le b\le1$ | Conditionally convergent |
| otherwise ($|a|\ge\max(b,1)$) | Diverges |
Additional Notes
- The threshold $\max(b,1)$ elegantly unifies two competing growth mechanisms: for $b>1$ the geometric growth of $b^n$ dominates the linear $n$, while for $b\le1$ the roles reverse and $n$ alone sets the scale — a clean illustration of how asymptotic dominance shifts the effective comparison series.
- This problem generalizes both a pure geometric series (formally $b\to\infty$, giving threshold $|a|<b$) and the alternating harmonic-type series of Problem 108 (the case $b=0$, $a=-1$), unifying two previously separate convergence mechanisms under one parametrized family.
- Only a single boundary value of $a$ (namely $a=-1$, and only when $0\le b\le1$) rescues convergence at the threshold — every other point on the boundary circle $|a|=\max(b,1)$ diverges, making this a relatively "thin" set of conditionally convergent parameter values compared to problems like 132 or 133 where an entire continuum of exponents supports conditional convergence.
Result
Convergence depends on $a$ and $b$ as follows: