Using Dirichlet's test, prove the convergence of: sum
Using Dirichlet's test, prove the convergence of: $$ \sum_{n=1}^{\infty} \frac{(-1)^{\frac{n(n-1)}{2}}}{n}. $$
Explanation of Method
Dirichlet's test needs the partial sums of $a_n=(-1)^{n(n-1)/2}$ to stay bounded while $b_n=1/n$ decreases to $0$. Since $n(n-1)/2\bmod2$ cycles through a period-4 pattern $(0,1,1,0)\pmod2$, i.e. $a_n$ cycles through $(1,-1,-1,1)$ which sums to zero over each full period, the partial sums $A_N$ never exceed $1$ in absolute value.
- Determine the period-4 sign pattern of $a_n=(-1)^{n(n-1)/2}$.
- Show the partial sums $A_N$ are bounded by $1$.
- Apply Dirichlet's test with $b_n=1/n$.
Step 1 · Determine the sign pattern
Computing $\frac{n(n-1)}2\bmod2$ for $n=1,\ldots,8$ gives the values $0,1,3,6,10,15,21,28\pmod2 = 0,1,1,0,0,1,1,0$, so $a_n=(-1)^{n(n-1)/2}$ cycles with period $4$: $1,-1,-1,1,1,-1,-1,1,\ldots$
Step 2 · Bound the partial sums
Since each full period sums to $1-1-1+1=0$: $A_{4k}=0$, $A_{4k+1}=1$, $A_{4k+2}=0$, $A_{4k+3}=-1$. So $|A_N|\le1$ for every $N$ — the partial sums of $a_n$ are bounded.
Step 3 · Apply Dirichlet's test
With $b_n=\frac1n$: $b_n$ is monotonically decreasing and $b_n\to0$. All three conditions of Dirichlet's test hold ($|A_N|\le1$ bounded, $b_n$ decreasing, $b_n\to0$), so the series $\sum a_nb_n$ converges.
Summary Table
| Quantity | Value |
|---|---|
| Sign pattern of $a_n$ (period 4) | $1,-1,-1,1$ |
| Bound on $|A_N|$ | $1$ |
| Conclusion | Converges |
Additional Notes
- Written out, the series is $\frac11-\frac12-\frac13+\frac14+\frac15-\frac16-\frac17+\frac18+\cdots$ — a "double alternation" pattern (two same signs, then two opposite) rather than the simple alternation of the classical alternating harmonic series, yet Dirichlet's test handles it with equal ease since only boundedness of partial sums matters, not any particular periodicity length.
- This is a template for handling any series where the sign sequence is periodic with mean zero over each period: as long as the period sums to zero, the partial sums stay bounded (by at most the largest partial sum within one period), letting Dirichlet's test apply regardless of the period's length or internal structure.
- Problem 113 uses the same Dirichlet-test machinery but with a far less regular sign sequence — one governed by $[\sqrt n]$ rather than a fixed period — showing the technique's robustness to increasingly irregular (but still boundedness-preserving) sign patterns.
Result
The series converges by Dirichlet's test: