Using Dirichlet's test, prove the convergence of: sum
Using Dirichlet's test, prove the convergence of: $$ \sum_{n=1}^{\infty} \frac{\ln^2 n}{n} \sin \frac{n\pi}{4}. $$
Explanation of Method
Unlike the previous Dirichlet-test problems, the coefficient $a_n=\frac{\ln^2n}n$ is not monotonic on all of $n\ge1$ — it first increases, then decreases once $n$ exceeds $e^2\approx7.39$. Splitting off the finite initial segment where monotonicity fails leaves a tail that is genuinely decreasing to $0$, and Dirichlet's test applies to that tail exactly as usual, with $\sin\frac{n\pi}4$ supplying bounded (period-8) partial sums.
- Determine where $a_n=\frac{\ln^2n}n$ is monotonically decreasing.
- Verify $a_n\to0$ and bound the partial sums of $\sin\frac{n\pi}4$.
- Apply Dirichlet's test to the tail, and combine with the finite initial segment.
Step 1 · Determine monotonicity of $a_n$
With $f(x)=\dfrac{\ln^2x}x$: $f'(x)=\dfrac{\ln x(2-\ln x)}{x^2}$, which is negative exactly when $\ln x>2$, i.e. $x>e^2\approx7.39$. So $a_n$ increases for $n\le7$ and strictly decreases for $n\ge8$.
Step 2 · Verify the limit and bound $\sin$'s partial sums
By L'Hôpital's rule (applied twice), $\displaystyle\lim_{n\to\infty}\frac{\ln^2n}n=0$. And $\sin\frac{n\pi}4$ is periodic with period $8$, taking the repeating values $\frac{\sqrt2}2,1,\frac{\sqrt2}2,0,-\frac{\sqrt2}2,-1,-\frac{\sqrt2}2,0$, so using the closed-form sine-sum identity with $\theta=\pi/4$:
$$ \left|\sum_{n=1}^N \sin\frac{n\pi}4\right| \le \frac1{\sin(\pi/8)} \approx 2.61. $$
Step 3 · Apply Dirichlet's test to the tail
Split $\sum_{n=1}^{\infty} = \sum_{n=1}^{7} + \sum_{n=8}^{\infty}$. The first sum is a finite constant. For $n\ge8$, $a_n=\frac{\ln^2n}n$ is decreasing with $a_n\to0$, and the partial sums of $\sin\frac{n\pi}4$ are bounded (Step 2), so Dirichlet's test applies to the tail, giving convergence of $\sum_{n=8}^{\infty}a_n\sin\frac{n\pi}4$. Adding back the finite initial sum, the full series converges.
Summary Table
| Quantity | Value |
|---|---|
| $a_n$ monotonic decreasing for | $n\ge8$ |
| $\lim a_n$ | $0$ |
| Bound on $\left|\sum\sin\frac{n\pi}4\right|$ | $\approx2.61$ |
| Conclusion | Converges |
Additional Notes
- This is a good reminder that Dirichlet's test only truly requires "eventual" monotonicity — since any finite number of initial terms can always be split off and summed separately without affecting convergence, non-monotonic behavior on a bounded initial segment never threatens the overall conclusion.
- The specific threshold $x=e^2$ where monotonicity begins is a direct consequence of differentiating $\ln^2x/x$, and this exact "$\ln^2x$ increases-then-decreases" pattern recurs whenever a power of $\ln n$ competes against a linear factor in the denominator.
- The period-8 structure of $\sin\frac{n\pi}4$ here mirrors the period-4 structure of Problem 112's $(-1)^{n(n-1)/2}$: both are examples where the "oscillating" sequence has an exact finite period rather than the more general closed-form bound needed in Problems 114 and 115.
Result
The series converges by Dirichlet's test applied to the tail $n\ge8$: