Using the Cauchy criterion, prove that the convergence
Using the Cauchy criterion, prove that the convergence of the series $\sum_{n=1}^{\infty} a_n$ implies the convergence of the series $\sum_{n=1}^{\infty} A_n$, where $$ A_n = \sum_{i=p_n}^{p_{n+1}-1} a_i \qquad (p_1=1,\; p_n < p_{n+1},\; n=1,2,\ldots). $$
Explanation of Method
Each $A_n$ is just a block of consecutive terms of $a_n$, so the partial sums of $\sum A_n$ are a subsequence of the partial sums of $\sum a_n$ (specifically, the tail sums $\sum_{j=N+1}^{M} A_j$ collapse to a single block $\sum_{i=p_{N+1}}^{p_{M+1}-1} a_i$ of the original series). Since $\sum a_n$ converges, its Cauchy tail condition applies to any such block once the starting index is large enough — and that is exactly the Cauchy condition needed for $\sum A_n$.
- State the Cauchy criterion for series convergence.
- Show that any block of consecutive $A_j$'s telescopes into a single block of the original $a_i$'s.
- Transfer the Cauchy bound from $\sum a_n$ to $\sum A_n$.
Step 1 · State the Cauchy criterion
$\sum_{n=1}^{\infty} a_n$ converges if and only if for every $\varepsilon>0$ there exists $N_0$ such that for all $m>n\ge N_0$:
$$ \left|\sum_{k=n+1}^{m} a_k\right| < \varepsilon. $$
Step 2 · Understand the grouping and relate it to the original series
The sequence $\{p_n\}$ partitions the positive integers, and $A_n = a_{p_n}+a_{p_n+1}+\cdots+a_{p_{n+1}-1}$. For any $M>N$:
$$ \sum_{j=N+1}^{M} A_j = \sum_{j=N+1}^{M}\sum_{i=p_j}^{p_{j+1}-1} a_i = \sum_{i=p_{N+1}}^{p_{M+1}-1} a_i, $$
since the blocks $A_{N+1},\ldots,A_M$ are consecutive and non-overlapping, so their union is exactly the range $[p_{N+1}, p_{M+1}-1]$.
Step 3 · Apply the Cauchy criterion to the grouped series
Let $\varepsilon>0$. Since $\sum a_n$ converges, choose $N_0$ as above. Since $p_n\to\infty$, choose $N$ with $p_N\ge N_0$. Then for any $M>N$, since $p_{N+1}\ge p_N\ge N_0$ and $p_{M+1}-1\ge p_{N+1}$:
$$ \left|\sum_{j=N+1}^{M} A_j\right| = \left|\sum_{i=p_{N+1}}^{p_{M+1}-1} a_i\right| < \varepsilon. $$
This is precisely the Cauchy criterion for $\sum_{n=1}^{\infty} A_n$, so this series converges.
Summary Table
| Object | Meaning |
|---|---|
| $A_n$ | Block sum $\sum_{i=p_n}^{p_{n+1}-1} a_i$ |
| $\sum_{j=N+1}^{M} A_j$ | Equals $\sum_{i=p_{N+1}}^{p_{M+1}-1} a_i$ |
| Conclusion | $\sum a_n$ convergent $\implies$ $\sum A_n$ convergent |
Additional Notes
- This result is known as the Grouping Theorem, or associativity of convergent series: inserting parentheses into a convergent series (without reordering terms) never breaks convergence.
- The converse fails: grouping can turn a divergent series convergent. For example, $\sum(-1)^{n-1}=1-1+1-1+\cdots$ diverges, but grouping in pairs $(1-1)+(1-1)+\cdots$ gives the convergent series $\sum 0 = 0$.
- This is a purely structural fact about partial sums — it works for any partition $\{p_n\}$ of the indices, regardless of what the terms $a_n$ actually are, as long as the original series converges.
Result
If $\sum_{n=1}^{\infty} a_n$ converges, then: