Show that, in the setting of the previous problem: sum

Show that, in the setting of the previous problem: $$ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} A_n, $$ where $A_n = \sum_{i=p_n}^{p_{n+1}-1} a_i$ with $p_1=1$, $p_n

Explanation of Method

Problem 21 established that $\sum A_n$ converges once $\sum a_n$ does; now the actual value of that sum needs to be pinned down. The key observation is that the $n$-th partial sum $T_n$ of the grouped series is exactly equal to a specific partial sum $S_{p_{n+1}-1}$ of the original series — no approximation, an exact equality. Since $p_{n+1}-1\to\infty$, this identifies $\{T_n\}$ as a subsequence of $\{S_m\}$, and subsequences of a convergent sequence inherit its limit.

Express the partial sum $T_n$ of the grouped series in terms of the original partial sums $S_m$.
Recognize $\{T_n\}$ as a subsequence of $\{S_m\}$.
Conclude equality of the limits using the subsequence property of convergent sequences.

Step 1 · Define partial sums

Let $S_m=\sum_{k=1}^{m}a_k$ and $T_n=\sum_{j=1}^{n}A_j$, so that $\sum a_n=\lim_{m\to\infty}S_m$ and $\sum A_n=\lim_{n\to\infty}T_n$.

Step 2 · Express $T_n$ in terms of the original series

Expanding and using that the blocks are consecutive and non-overlapping:

$$ T_n = \sum_{j=1}^{n} A_j = \sum_{i=p_1}^{p_2-1}a_i + \sum_{i=p_2}^{p_3-1}a_i + \cdots + \sum_{i=p_n}^{p_{n+1}-1}a_i = \sum_{i=p_1}^{p_{n+1}-1} a_i. $$

Since $p_1=1$:

$$ T_n = \sum_{i=1}^{p_{n+1}-1} a_i = S_{p_{n+1}-1}. $$

Step 3 · Relate the limits via a subsequence

Since $\{p_n\}$ is strictly increasing with $p_1=1$, $p_n\to\infty$, so $m_n:=p_{n+1}-1\to\infty$ as well; $\{S_{m_n}\}$ is therefore a subsequence of $\{S_m\}$. Because $\sum a_n$ converges, $\{S_m\}$ converges, and every subsequence of a convergent sequence converges to the same limit:

$$ \sum_{n=1}^{\infty} A_n = \lim_{n\to\infty} T_n = \lim_{n\to\infty} S_{p_{n+1}-1} = \lim_{m\to\infty} S_m = \sum_{n=1}^{\infty} a_n. $$

Summary Table

Quantity	Relation
$T_n$ (partial sum of $\sum A_n$)	$=S_{p_{n+1}-1}$
$\{S_{p_{n+1}-1}\}$	Subsequence of $\{S_m\}$
Conclusion	$\sum a_n = \sum A_n$

Additional Notes

This is the precise sense in which grouping is "just adding parentheses": the exact equality $T_n = S_{p_{n+1}-1}$ shows no terms are added, removed, or reordered — the grouped partial sums are literally a subset of the original partial sums.
Together, Problems 21 and 22 form the full associativity statement for series: convergence is preserved, and the sum is unchanged, whenever terms of a convergent series are grouped by inserting parentheses without reordering.
This associativity should not be confused with commutativity (rearrangement): the Riemann Rearrangement Theorem shows that reordering the terms of a conditionally convergent series can change its sum entirely, which is a fundamentally different (and much more delicate) operation than grouping.

Result

Grouping the terms of a convergent series leaves its sum unchanged:

$$ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} A_n. $$