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!


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 $p$.

  • $\begingroup$ 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. $\endgroup$
    – YiFan Tey
    Sep 19, 2019 at 1:22
  • 7
    $\begingroup$ @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. $\endgroup$ Sep 19, 2019 at 2:13
  • 3
    $\begingroup$ 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 $\endgroup$ Sep 19, 2019 at 2:16
  • 10
    $\begingroup$ 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. $\endgroup$ Sep 19, 2019 at 2:23
  • $\begingroup$ @Jack: Ah, I see, I was being naïve. $\endgroup$
    – YiFan Tey
    Sep 19, 2019 at 2:27

2 Answers 2


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 $a \in X$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .