TAOCP 1.2.3 Exercise 7

Section 1.2.3: Sums and Products

Exercise 7. [HM23] Given that $c$ is an integer, show that $\sum_{R(j)} a_j = \sum_{R(c-j)} a_{c-j}$, even if both series are infinite.

Verified: yes
Solve time: 37s

Solution

Let

$S=\sum_{R(j)} a_j.$

Define the change of variable

$i=c-j.$

Since $c$ is an integer, the mapping

$j \mapsto c-j$

is a permutation of the integers. Indeed, for every integer $i$ there exists exactly one integer $j=c-i$ such that

$i=c-j.$

Equation (5) states that

$\sum_{R(j)} a_j=\sum_{R(p(j))} a_{p(j)}$

whenever $p(j)$ is a permutation of the relevant values. Taking

$p(j)=c-j,$

we obtain

$\sum_{R(j)} a_j=\sum_{R(c-j)} a_{c-j}.$

The text following Eq. (5) states explicitly that the replacement of $j$ by $p(j)$ is always valid for infinite sums when

$p(j)=c\pm j.$

Hence the transformation above remains valid even if both series are infinite.

Therefore

$\boxed{\sum_{R(j)} a_j=\sum_{R(c-j)} a_{c-j}}.$

This completes the proof.

∎