OFFSET
3,2
COMMENTS
With summation by parts to obtain 1st formula:
Sum_{k>=1} (1/length(k)^length(k)) =
Sum_{m=1..9} (1/1^1) + Sum_{m=10..99} (1/2^2) + Sum_{m=100...999} (1/3^3) + Sum_{m=1000...9999} (1/4^4) + ... =
9*(1/1^1) + 90*(1/2^2) + 900*(1/3^3) + 9000*(1/4^4) + 90000*(1/5^5) + ... =
9 ( 1/1^1 + 10^1/2^2 + 10^2/3^3 + 10^3/4^4 + 10^4/5^5 + ... =
(9/10) * (10^1/1^1 + 10^2/2^2 + 10^3/3^3 + 10^4/4^4 + 10^5/5^5 + ... =
(9/10) * ( (10/1)^1 + (10/2)^2 + (10/3)^3 + (10/4)^4 + (10/5)^5 + ... =
(9/10) * Sum_{m>=1} (10/m)^m.
REFERENCES
Xavier Merlin, Methodix Analyse, Ellipses, 1997, Exercice 22 p. 120.
J.-M. Monier, Analyse, Tome 3, 2ème année, MP.PSI.PC.PT, Dunod, 1997, Exercice 3.2.1.h" p. 248.
FORMULA
Equals (9/10) * Sum_{k>=1} (10/k)^k.
Equals Sum_{n>=1} (1/A138908(n)).
EXAMPLE
168.05245375262168949085673320556724...
MAPLE
evalf((9/10) * Sum((10/n)^n, n=1..infinity), 100);
PROG
(PARI) (9/10) * suminf(k=1, (10/k)^k) \\ Michel Marcus, Jun 08 2019
CROSSREFS
KEYWORD
AUTHOR
Bernard Schott, May 19 2019
STATUS
approved