TAOCP 1.2.3 Exercise 3
The first sum is $\sum_{0 \le n \le 5} \frac{1}{2n+1}.$ The integers satisfying $0 \le n \le 5$ are $n=0,1,2,3,4,5.$ Hence
Section 1.2.3: Sums and Products
Exercise 3. ▶ [13] Without using the $\sum$-notation, write out the equivalent of
$$\sum_{0 \le n \le 5} \frac{1}{2n + 1},$$
and also the equivalent of
$$\sum_{0 \le n^2 \le 5} \frac{1}{2n^2 + 1}.$$
Explain why the two results are different, in spite of rule (b).
Verified: yes
Solve time: 44s
Solution
The first sum is
$\sum_{0 \le n \le 5} \frac{1}{2n+1}.$
The integers satisfying $0 \le n \le 5$ are
$n=0,1,2,3,4,5.$
Hence
$$ \sum_{0 \le n \le 5} \frac{1}{2n+1} = \frac11+\frac13+\frac15+\frac17+\frac19+\frac1{11}. $$
The second sum is
$\sum_{0 \le n^2 \le 5} \frac{1}{2n^2+1}.$
The condition $0 \le n^2 \le 5$ means that $n^2$ can equal only $0,1,4$, since $2^2=4$ and $3^2=9>5$. Therefore the integers satisfying the condition are
$n=-2,-1,0,1,2.$
Thus
$$ \sum_{0 \le n^2 \le 5} \frac{1}{2n^2+1} = \frac19+\frac13+1+\frac13+\frac19. $$
The two results are different because rule (b) applies only when the change of variable is a permutation of the relevant values. The transformation $n \mapsto n^2$ is not a permutation on the integers satisfying $0 \le n \le 5$, since
$$$$
Different values of $n$ produce the same value of $n^2$, so terms are repeated in the second sum. Therefore rule (b) does not justify replacing $n$ by $n^2$ in this case.
The answers are
$$ \boxed{ \sum_{0 \le n \le 5} \frac{1}{2n+1} = 1+\frac13+\frac15+\frac17+\frac19+\frac1{11} } $$
and
$$ \boxed{ \sum_{0 \le n^2 \le 5} \frac{1}{2n^2+1} = \frac19+\frac13+1+\frac13+\frac19 }. $$