Using the Cauchy criterion, prove that the series sum
Using the Cauchy criterion, prove that the series $\displaystyle\sum_{n=3}^{\infty} \frac{1}{n}$ diverges.
Explanation of Method
To disprove the Cauchy criterion, it suffices to exhibit one fixed $\varepsilon_0>0$ for which the tail sum never shrinks below $\varepsilon_0$, no matter how far out the tail starts. Grouping the harmonic terms from $N+1$ to $2N$ — exactly $N$ terms, each at least as large as the smallest one, $\frac1{2N}$ — gives a sum of at least $\frac12$ for every choice of $N$, which is the classic Oresme argument for divergence of the harmonic series.
- State the negation of the Cauchy criterion (what divergence requires).
- Construct a block of $N$ terms, from $N+1$ to $2N$, and bound it below.
- Conclude that the Cauchy criterion fails, so the series diverges.
Step 1 · Negation of the Cauchy criterion
A series diverges if there exists $\varepsilon_0>0$ such that for every $N$, some $m>n\ge N$ gives $\left|\sum_{k=n+1}^{m}a_k\right|\ge\varepsilon_0$. Here all terms $\frac1k$ are positive, so $\left|\sum_{k=n+1}^{m}\frac1k\right|=\sum_{k=n+1}^{m}\frac1k$.
Step 2 · Construct the block and bound it below
Let $N\ge3$ be arbitrary. Take $n=N$, $m=2N$:
$$ \sum_{k=N+1}^{2N} \frac1k. $$
Each of the $N$ terms in this sum satisfies $\frac1k \ge \frac1{2N}$ (since $k\le2N$), so:
$$ \sum_{k=N+1}^{2N}\frac1k \ge N\cdot\frac1{2N} = \frac12. $$
Step 3 · Apply the negation of the Cauchy criterion
Taking $\varepsilon_0=\frac12$: for every $N\ge3$, choosing $m=2N>n=N\ge N$ gives $\sum_{k=N+1}^{2N}\frac1k \ge \frac12=\varepsilon_0$. So the Cauchy criterion fails, and $\sum_{n=3}^{\infty}\frac1n$ diverges.
Summary Table
| Quantity | Value |
|---|---|
| Block chosen | $k=N+1,\ldots,2N$ ($N$ terms) |
| Smallest term in block | $\dfrac1{2N}$ |
| Block sum lower bound | $\ge\dfrac12$ for every $N$ |
| Conclusion | Diverges (Cauchy criterion fails) |
Additional Notes
- Removing the first two terms ($n=1,2$) from the harmonic series does not affect divergence — dropping finitely many terms from any series never changes whether it converges or diverges, only the value of the sum if it does converge.
- This is the classical medieval argument (attributed to Nicole Oresme) for divergence of the harmonic series, and it generalizes: grouping in blocks that double in length each time ($[2^k+1, 2^{k+1}]$) shows the partial sums grow at least linearly in $k$, hence without bound.
- An equivalent grouping view: $\frac13+\frac14+\frac15+\cdots \ge \frac14+\frac14+\frac16+\frac16+\cdots$, replacing each term with the smallest one in a pair or group, again producing a divergent lower bound.
Result
The series diverges, since the Cauchy criterion fails with $\varepsilon_0=\tfrac12$: