Closed form for $S_n = \sum_{i = 0}^n r^{n^i} = r + r^n + r^{n^2} + r^{n^3} + \cdots + r^{n^n}$?

Asked Aug 17 '17 at 00:01

Active Aug 17 '17 at 00:15

Viewed 111 times

Find a closed form equation for the sum $$S_n = \sum_{i = 0}^n r^{n^i} = r + r^n + r^{n^2} + r^{n^3} + \cdots + r^{n^n}$$

I don't really think this is solvable. I tried with multiplying both side by something, and writing an terms of $S_n$ but it didn't help. Is it actually solvable?

edited Aug 17 '17 at 00:15

Thomas Andrews

177,126

asked Aug 17 '17 at 00:01

Rezwan Arefin

3,106
1
14
39

1

The limit as $n\to\infty$ is a Lacunary functions, which have no such closed form. If there were a closed form in terms of $n$, then there'd probably be a closed form for the limit. – Simply Beautiful Art Aug 17 '17 at 00:05
@SimplyBeautifulArt I didn't know about Lacunary Functions. But all the functions in the wiki page has sum to the infinity .. My one in finite :| (lol) – Rezwan Arefin Aug 17 '17 at 00:10
@SimplyBeautifulArt Actually, given the repetition of $n$ in the sum makes this not a series, and thus not related to Lacunary. Indeed, $S_n\to r$ for $|r|<1$. – Thomas Andrews Aug 17 '17 at 00:12
Lacunary would be the limit of $\sum_{i=0}^{n} r^{q^i}$ for fixed $q$. @SimplyBeautifulArt – Thomas Andrews Aug 17 '17 at 00:13
Ah, okay @ThomasAndrews – Simply Beautiful Art Aug 17 '17 at 00:16
Well still, I highly doubt of any closed form. – Simply Beautiful Art Aug 17 '17 at 00:16
Me too. @SimplyBeautifulArt :) – Thomas Andrews Aug 17 '17 at 00:17
1

Why would you even imagine it has a closed form? – GEdgar Aug 17 '17 at 00:22
5

@GEdgar Why would you discourage curiosity? – Unit Aug 17 '17 at 00:24
See also https://math.stackexchange.com/q/1072907/24908 for the case when r = n = 2. – b_jonas Mar 22 '23 at 11:54

Closed form for $S_n = \sum_{i = 0}^n r^{n^i} = r + r^n + r^{n^2} + r^{n^3} + \cdots + r^{n^n}$?

0 Answers0

Linked