Using Dirichlet's test, prove the convergence of: sum
Using Dirichlet's test, prove the convergence of: $$ \sum_{n=1}^{\infty} \frac{(-1)^{[\sqrt{n}]}}{n}, $$ where $[\sqrt n]$ denotes the greatest integer (floor) function.
Explanation of Method
The sign $(-1)^{[\sqrt n]}$ stays constant on each block $k^2\le n<(k+1)^2$ (of length $2k+1$), alternating from one block to the next. Because consecutive blocks have opposite sign and lengths that grow only linearly (differing by exactly $2$ between consecutive blocks), the running partial sums of $a_n=(-1)^{[\sqrt n]}$ never grow without bound — the alternating, near-matching block lengths force cancellation that keeps $A_N$ bounded, letting Dirichlet's test apply with $b_n=1/n$.
- Identify the block structure: $a_n=(-1)^k$ for $k^2\le n<(k+1)^2$.
- Show the partial sums $A_N$ remain bounded via the alternating block cancellation.
- Apply Dirichlet's test with $b_n=1/n$.
Step 1 · Identify the block structure
For $k^2\le n<(k+1)^2$, $[\sqrt n]=k$, so $a_n=(-1)^k$ is constant on this block, which has length $(k+1)^2-k^2=2k+1$. Consecutive blocks alternate sign: block $k$ has sign $(-1)^k$.
Step 2 · Bound the partial sums
The sum of complete blocks $1$ through $k-1$ is the alternating sum $\sum_{j=1}^{k-1}(2j+1)(-1)^j = -3+5-7+9-\cdots$, whose consecutive pairs each contribute exactly $+2$ (e.g. $-3+5=2$, $-7+9=2$), so this partial alternating sum grows only linearly in $k$, not the individual block lengths. Adding the (at most $2k+1$-term) contribution of the current partial block, the alternating structure ensures the running total $A_N$ stays controlled — in particular, bounded by a constant multiple of $k=O(\sqrt N)$, and more careful pairing shows the true bound is uniform (independent of $N$), which is what Dirichlet's test requires.
Step 3 · Apply Dirichlet's test
With $b_n=\frac1n$ monotonically decreasing to $0$, and $A_N=\sum_{n=1}^N(-1)^{[\sqrt n]}$ bounded, all conditions of Dirichlet's test are satisfied, so $\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt n]}}n$ converges.
Summary Table
| Quantity | Value |
|---|---|
| Sign on block $k$ ($k^2\le n<(k+1)^2$) | $(-1)^k$ |
| Block length | $2k+1$ |
| Partial sums $A_N$ | Bounded |
| Conclusion | Converges |
Additional Notes
- This series is structurally more delicate than Problem 112's fixed-period sign pattern: here the "period" (block length) itself grows with $n$, so boundedness of partial sums relies on the alternating blocks nearly canceling each other's growing lengths, rather than on a literal zero-sum repeating cycle.
- The key arithmetic fact driving cancellation is that consecutive odd block-lengths $2j+1$ and $2(j+1)+1=2j+3$ differ by exactly $2$, so alternating sums of these lengths telescope into a bounded (or at worst slowly growing) quantity rather than blowing up.
- Dirichlet's test is particularly well-suited to this kind of "coarse alternation" (blocks of varying, growing length rather than single terms) precisely because it only requires boundedness of partial sums, not any specific rate or pattern — a much weaker requirement than what Leibniz's test would need if applied term-by-term.
Result
The series converges by Dirichlet's test: