Show that if $$a_1 + a_2 + a_3 + ...$$

is a convergent series of real numbers and $v_1, v_2, v_3, ...$ is a subsequence of the sequence $1, 2, 3, ...$ then

$$(a_1 + a_2 + .. + a_{v_1}) + (a_{v_1 + 1} + .. + a_{v_2}) + (a_{v_2 + 1} + ... + a_{v_3}) + ... = \sum\limits_{n = 1}^{\infty} a_n$$ How would you show that this converges to the summation? I considered using sums of series.

  • $\begingroup$ I've edited your post to include Latex. Please verify that it's correct. $\endgroup$ – user61527 Oct 1 '13 at 4:10
  • $\begingroup$ Yes it is. Thank-you. $\endgroup$ – William Hsu Oct 1 '13 at 4:18

Hint: If $S_N = \sum_{n=1}^\infty a_n$ denotes the $N$-th partial sum of the original series, can you relate $S_{v_n}$ to a partial sum of the new series?

Edit: I've added a more-complete solution/sketch below, but I'd encourage you to try to think about how to do this using my hint before looking at it!

Solution: Set $v_0 = 0$ to make sigma notation easier. Prove that the $N$-th partial sum $T_N = \sum_{i=1}^N(\sum_{k=v_{i-1}+1}^{v_i}a_k)$ of your new series is equal to $S_{v_N}$, for each $N \geq 1$. Since $\sum_{n=1}^\infty a_n$ converges, by definition the sequence $\{S_N\}$ converges, and $\lim_{N\to\infty}S_N = \sum_{n=1}^\infty a_n$. Since every subsequence of a convergent sequence also converges to the same value, it follows that $\{T_N\} = \{S_{v_n}\}$ converges to $\sum_{n=1}^\infty a_n$. Hence $\sum_{i=1}^\infty(\sum_{k=v_{i-1}+1}^{v_i}a_k) = \sum_{n=1}^\infty a_n$.


Let $s_n = a_1+...+a_n$. Then by assumption $s_n \to s$. If $s_{n_k}$ is a subsequence, it is straightforward to show that $s_{n_k} \to s$ also.

Then let $b_1 = s_{v_1}$, and for $k>1$, let $b_k = s_{v_{k}} - s_{v_{k-1}}$, then $b_1,b_2,...$ are the terms in parentheses in the question.

The problem is to show that $b_1+\cdots + b_n$ is a convergent series. It is easy to see that $b_1+\cdots + b_n = s_{v_n}$. Since $s_{v_n} \to s$, we see that $\lim_n (b_1+\cdots + b_n) = s$.


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