# Changing order of sum

We know that $$\sum_{n=1}^{\infty}a_n = \mbox{convergent}$$ Does that imply that $$a_3 + a_1 + a_2 + a_6 + a_4 + a_5 + \cdots = \mbox{convergent?}$$ We don't know whether our series is absolutely convergent.

• To be clear: the pattern is $3,1,2,\;6,4,5,\;9,7,8,\dots$ and so on, right? – Omnomnomnom Dec 20 '15 at 23:51
• Little "local" changes of order make no difference. From the fact that partial sums of the $a_i$ converge, one can show that the modified partial sums converge to the same value. – André Nicolas Dec 20 '15 at 23:54
• Yes, the permutation has bounded displacements, see this previous question math.stackexchange.com/questions/1150964/… – orangeskid Dec 20 '15 at 23:56
• Hey I have a question, why is that series convergent? Doesn't it add up to infinity? – Aldon Dec 21 '15 at 0:10

Given that $$\sum_{k=1}^{\infty} a_k=\mbox{convergent}$$ We have $$a_3 + a_1 + a_2 + a_6 + a_4 + a_5 + \cdots$$ $$=\sum_{k=1}^{\infty} \left(a_{3k} +a_{3k-2}+a_{3k-1}\right)$$ $$=\sum_{k=1}^{\infty} \left(a_{3k-2}+a_{3k-1}+a_{3k}\right)$$ $$=a_1 + a_2 + a_3 + a_4 + a_5 + a_6+\cdots$$ $$=\sum_{k=1}^{\infty} a_k =\mbox{convergent}$$

Let $S_n$ be the nth partial sum of the original series, and let $T_n$ be the nth partial sum of the new series.

Since the original series converges, $S_n\to S$ for some number $S$.

Since $T_{3n}=S_{3n}, \;T_{3n+1}=S_{3n}+a_{3n+3},\; T_{3n+2}=S_{3n}+a_{3n+3}+a_{3n+1}$, and $a_n\to 0$,

$T_n\to S$ so the new series converges to the same sum.

Let $b_n$ denote the $n$th element of your sequence (so for example $b_1 = a_3$). Let $S_N = \sum_{n=1}^N a_n$ and $T_N = \sum_{n=1}^N b_n$.
We note that $T_{3N} = S_{3N}$. So, if $T_N$ converges, it converges to the same limit as $S_N$.
Moreover, $T_{3N+2} = T_{3N} + b_{3N+1} + b_{3N+2}$. However, we note that $b_n \to 0$ by the convergence of the original sum. So, for any $\epsilon > 0$, we can select $N_0$ so that $b_N < \epsilon$ when $N > N_0$. So, we have $$|T_{3N + 2} - T_{3N}| = |b_{3N+1} + b_{3N+2}| < 2 \epsilon$$ this is enough to conclude that $T_N$ is Cauchy, and thus convergent.
From the fact that partial sums of the $a_i$ converge, say to $s$, one can show that the modified partial sums converge to $s$. Let $(b_n)$ be our modified sequence. Then $\sum_1^{3k} a_i =\sum_1^{3k}b_i$, and both have limit $s$.
Since the $a_i$ have limit $0$, we have that $\lim \sum_1^{3k+1} b_i$ and $\lim \sum_1^{3k+2} b_i$ are also equal to $s$.