Change of variables for summation fails

Question

Suppose I have the sum $ \displaystyle S_1 = \sum_{i=1}^5 \cfrac{1}{(i+1)^2} $

I can make a change of variables by letting $ j=i+1 $. When $ i=1 \implies j=2 $, and when $ i=5 \implies j=6 $

$ \displaystyle \therefore S_1 = \sum_{j=2}^6 \cfrac{1}{j^2} $.

But if I were to make a change of variables with $ j=(i+1)^2 $ and use the same logic as above, the resulting sum would be: $ \displaystyle \sum_{j=4}^{36}\cfrac{1}{j} $. Which is not equal to the original sum $ S_1 $.

In theory, why does this change of variable fail?

Answer 1

The main idea is that you don't get all values from $4$ to $36$ in $j=(i+1)^2$, since not all values from $4$ to $36$ can be represented this way when $i\in\{1,2,3,4,5\}$. An easy way to check this is that the substitution that you applied is one-to-one, but the original summation has $5$ terms whilst the modified summation has $33$. Here's an explanation:

When $j=i+1$, the possible values are $$i=1\implies j=2$$ $$i=2\implies j=3$$ $$i=3\implies j=4$$ $$i=4\implies j=5$$ $$i=5\implies j=6$$ So, indeed, your change of variables is correct. However, when $j=(i+1)^2$, we have $$i=1\implies j=4$$ $$i=2\implies j=9$$ $$i=3\implies j=16$$ $$i=4\implies j=25$$ $$i=5\implies j=36$$ So, instead of $$\sum_{j=4}^{36}\frac{1}{j}=\sum_{j\in\{4,5,6,7,\cdots, 36\}}\frac{1}{j}$$ It is $$\sum_{j\in\{4,9,16,25,36\}}\frac{1}{j}$$

Answer 2

$\displaystyle \sum_{j=4}^{36}\cfrac{1}{j}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{36}$ whereas $\displaystyle \sum_{j=2}^{6}\cfrac{1}{j^2}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$. So their sums are not matching. Instead you can write $$\displaystyle \sum_{j=2}^{6}\cfrac{1}{j^2}=\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$$. $\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$ means that it is the sum of all values $\frac{1}{j}$ where $j$ belongs to the set {$4,9,16,25,36$}. This implies that $\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$ becomes $\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ which is the required value.