Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \prod_{k=1}^n \frac{\ln(1+k)}{\ln(1+a+k)}, \qquad a>0. $$
Explanation of Method
Since $u_n$ is itself a running product, $u_{n+1}=u_n\cdot\frac{\ln(n+2)}{\ln(a+n+2)}$, so the ratio $u_n/u_{n+1}$ is simply the reciprocal of the newest factor. Approximating $\ln(a+n+2)\approx\ln(n+2)+\frac a{n+2}$ shows the correction term in Raabe's formula is of order $\frac1{n\ln n}$ rather than $\frac1n$ — far too weak to reach the threshold, so the Raabe limit collapses to $0$ and the series diverges for every $a>0$.
- Express the ratio $u_n/u_{n+1}$ as the reciprocal of the newest product factor.
- Approximate using $\ln(1+x)\approx x$ to show the correction is $O\!\left(\frac1{n\ln n}\right)$.
- Evaluate the Raabe limit $r=0<1$ and conclude divergence.
Step 1 · Express the ratio
Since $u_{n+1}=u_n\cdot\dfrac{\ln(n+2)}{\ln(a+n+2)}$:
$$ \frac{u_n}{u_{n+1}} = \frac{\ln(a+n+2)}{\ln(n+2)}. $$
Step 2 · Approximate for large $n$
Writing $\ln(a+n+2)=\ln(n+2)+\ln\!\left(1+\frac a{n+2}\right)\approx\ln(n+2)+\frac a{n+2}$:
$$ \frac{u_n}{u_{n+1}}-1 \approx \frac{a/(n+2)}{\ln(n+2)} \sim \frac{a}{n\ln n}. $$
Step 3 · Apply Raabe's test
$$ r = \lim_{n\to\infty} n\cdot\frac{a}{n\ln n} = \lim_{n\to\infty}\frac{a}{\ln n} = 0. $$
Since $r=0<1$, by Raabe's test the series diverges for every $a>0$. Directly, $\ln u_n=-\sum_{k=1}^n\ln\frac{\ln(1+a+k)}{\ln(1+k)}\sim -a\sum\frac1{k\ln k}\to-\infty$ (the harmonic-type series $\sum\frac1{k\ln k}$ diverges), giving $u_n\sim C/(\ln n)^a$, and $\sum1/(\ln n)^a$ diverges for every $a>0$.
Summary Table
| Quantity | Value |
|---|---|
| $u_n/u_{n+1}-1$ | $\sim\dfrac a{n\ln n}$ |
| Raabe limit $r$ | $0$ |
| Asymptotic $u_n$ | $\sim C/(\ln n)^a$ |
| Conclusion | Diverges for all $a>0$ |
Additional Notes
- The Raabe limit of exactly $0$ (rather than a finite value less than $1$) reflects that the correction term $a/(n\ln n)$ decays strictly faster than $1/n$ — an even weaker growth rate than a typical divergent case, driven by the extra logarithmic factor $\ln n$ in the denominator.
- The final asymptotic $u_n\sim C/(\ln n)^a$ shows the terms decay to zero, but only logarithmically slowly — this is the same qualitative failure mode seen in Problem 42's iterated-logarithm series, where terms shrinking slower than any power of $n$ (even though they do shrink) is characteristic of divergence.
- No finite value of $a$ can rescue convergence here, in sharp contrast to Problems 76, 77, 84, and similar problems where a parameter threshold separates convergence from divergence — the logarithmic decay is simply too weak at every positive $a$.
Result
The series diverges for every $a>0$: