<s>[INST] Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
    \frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
    \frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
    (a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
    (a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are

*

*Are there any other simple nontrivial examples?  I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.

*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream?  Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?

 [/INST] It may qualify as trivial, but note that there are infinitely many such series.
Consider a divergent series with positive decreasing terms $\sum_{n=1}^\infty u_n$, such that $\sum_{n=1}^\infty u_n^2$ is convergent to $L$, then it's a known theorem that we can change the signs of $u_n$ so as to get a series convergent to any given real number. Just pick $\sqrt{L}$.
That is, there exist $\sigma_n\in\{+1,-1\}$ such that $\sqrt{L}=\sum_{n=1}^\infty \sigma_nu_n$ while $L=\sum_{n=1}^\infty (\sigma_nu_n)^2$.

The idea of the proof of the aforementioned theorem, assuming $x>0$ (otherwise start with negative terms): sum $u_n$ until you get a value greater than $x$, call it $v_1$. Then sum $-u_n$ until you get a value smaller than $x$, call the sum of the new terms $v_2$, and continue this process to define $v_k$ for all $k$. Then $\sum_k v_k$ is alternating and convergent to $x$.
</s>