# Sequences where $\sum\limits_{n=k}^{\infty}{a_n}=\sum\limits_{n=k}^{\infty}{a_n^2}$

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 $$p$$.

• Well, clearly if all the terms in the series are positive this cannot happen as one sum is strictly greater than the other with a difference which does not tend to zero. Sep 19, 2019 at 1:22
• @YiFan this is not true since the existence of terms greater than 1 will provide a buffer zone for the terms less than 1 to "leak" the error into. You may be thinking of the case where all the terms are less than 1. Sep 19, 2019 at 2:13
• For example, consider the series Σ_{k=1}^∞ (3/2^k). The sum is 3, and you can verify Σ_{k=1}^∞ (9/4^k) = 3 Sep 19, 2019 at 2:16
• If $\sum c_n=A$ and $\sum c_n^2=B$ then $\sum a_n=\sum a_n^2=A^2/B$ for $a_n=(A/B)c_n$. But maybe that's cheating. Sep 19, 2019 at 2:23
• @Jack: Ah, I see, I was being naïve. Sep 19, 2019 at 2:27

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 $$\sum_{n=k}^{\infty}{a_n}(1-a_n) = 0$$

This is one way of looking at these sequences. As your starting $$k$$ might not be $$0$$, we're inside a linear subspace of $$l^2$$, the space of sequences $$(a_0, a_1, \ldots)$$ whose sum $$\sum_{n=0}^{\infty} |a_n|^2$$ converges. As we also need the sum $$\sum_{n=0}^{\infty} a_n$$ to converge, that places us inside a somewhat smaller linear subspace $$V$$ of $$l^2$$.

In there, we're looking for sequences $$a = (a_0, a_1, \ldots)$$ such that $$f(a) = 0$$, where $$f(a) = \sum_{n=k}^{\infty} a_n - \sum_{n=k}^{\infty} a_n^2$$. This is a smooth function on $$V$$, and its differential at $$a$$ in the direction of $$b$$ is $$d_b f(a) = \sum_{n=k}^{\infty} b_n - 2 \sum_{n=k}^{\infty} a_n b_n.$$ If $$a$$ is the zero sequence in $$V$$ (that is, any sequence $$a$$ with $$a_n = 0$$ for $$n \geq k$$), then $$df(a) = 0$$. If $$a$$ is not the zero sequence, then there exists a $$b$$ such that $$d_bf(a) \not= 0$$, take for example $$b = -\frac12 a$$.

Thus $$f$$ is a smooth function whose differential is nondegenerate away from the linear subspace $$N$$ of $$V$$ defined by the zero sequences in $$V$$, and degenerate on that subspace. This means that the set $$X = \{a \in V \mid f(a) = 0 \}$$ of sequences that satisfy your identity is an infinite-dimensional hypersurface that is nonsingular outside of $$N$$, and has singularities on that set.

Gerry's comment shows that there is also a smooth projection function $$p : V \setminus N \to X \setminus N$$, that is, a smooth function $$p$$ such that $$p(a) = a$$ for any sequence $$a \in X$$.