Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty} \lambda^{\frac{a \log_\lambda n+b}{c \log_\lambda n+d}}, \qquad \lambda>1. $$
Explanation of Method
Substituting $m=\log_\lambda n$ turns the exponent into the rational function $\frac{am+b}{cm+d}$, which tends to $\frac ac$ as $n\to\infty$ (equivalently $m\to\infty$). So the general term behaves asymptotically like the constant $\lambda^{a/c}$ times a slowly-varying correction — meaning the series essentially reduces to comparing $\lambda^{a/c}$ against $1$, exactly as with a geometric series.
- Substitute $m=\log_\lambda n$ to rewrite the exponent as $\frac{am+b}{cm+d}$.
- Evaluate the limiting exponent $\frac ac$ as $m\to\infty$.
- Apply the convergence criterion based on the sign of $\frac ac$.
Step 1 · Substitute and simplify the exponent
With $m=\log_\lambda n$, the exponent becomes $\dfrac{am+b}{cm+d}$.
Step 2 · Evaluate the limiting exponent
As $n\to\infty$ (so $m\to\infty$):
$$ \frac{am+b}{cm+d} = \frac{a+\frac bm}{c+\frac dm} \longrightarrow \frac ac. $$
So $u_n=\lambda^{\frac{am+b}{cm+d}} \to \lambda^{a/c}$ as a fixed limiting value, with the general term asymptotically comparable to this constant (up to a slowly-varying logarithmic correction).
Step 3 · Apply the convergence criterion
If $\frac ac<0$: since $\lambda>1$, $\lambda^{a/c}<1$, and more precisely the term decays like a negative power of $n$ (through the logarithmic correction), giving convergence. If $\frac ac>0$: the term tends to a value $\ge$ some constant $>0$ (or grows), so $u_n\not\to0$ and the series diverges. If $a=0$: finer analysis of $b,c,d$ is needed.
Summary Table
| Condition | Conclusion |
|---|---|
| $a/c<0$ | Converges |
| $a/c>0$ | Diverges |
| $a=0$ | Depends on $b,c,d$ (finer analysis needed) |
Additional Notes
- This problem shows how a change of variables (here $m=\log_\lambda n$) can convert an awkward mixed exponential-logarithmic expression into a simple rational function, a technique reminiscent of the substitution $u=\ln x$ used throughout the integral-test problems (94–96).
- The parameters $b$ and $d$ only affect lower-order corrections to the exponent, not the leading asymptotic behavior — this mirrors how, in Problem 75's $a^{(n^2+1)/(n+1)}$, the "+1" additive constants in numerator and denominator didn't change the leading exponent limit either.
- Since $\lambda>1$ is fixed throughout, the sign of $a/c$ (not the specific value of $\lambda$) is what fully controls convergence — much like how the sign of the leading coefficient controls the long-run behavior of a rational function.
Result
The series converges if and only if $a/c<0$ (for $c\ne0$):