I was recently looking at the series $\sum_{n=1}^{\infty}{\sin{n}\over{n}}$, for which the value quite cleanly comes out to be ${1\over2}(\pi-1)$, which is a rather cool closed form.
I then wondered what would happen to the value of the series if all the terms in the series were squared.
Turns out... nothing happens!
$\displaystyle\sum_{n=1}^{\infty}{\left({\sin{n}\over{n}}\right)}^2=\sum_{n=1}^{\infty}{\sin{n}\over{n}}={1\over2}(\pi-1)$.
This is a rather cool result, and I was wondering if there are any other simple series that share this property? Or, more generalized, series for which raising the terms to the power $m$ yields the same result as raising them to the power $p$.
This a long comment not an answer.
Notice that a similar property holds for the related integral
$$ \int_{0}^{\infty}\frac{\sin x}{x}dx = \int_{0}^{\infty}\bigg(\frac{\sin x}{x}\bigg)^2 dx = \frac{\pi}{2}. $$
This makes me wonder if such sequence has to do something with the property of orthogonality. Two functions $f$ and $g$ are said to be orthogonal with weight $1$ $$ \int_{0}^{\infty}f(x)g(x) = 0 $$
If we impose the condition that $g(x) = 1 - f(x)$ then we are looking at a special case of orthogonality where $$ \int_{0}^{\infty}f(x)(1-f(x)) = 0 $$
which is analogous to OP question since $\sum_{n=k}^{\infty}{a_n}(1-a_n) = 0$