MathDB
TOT 027 1982 Autumn S1 sum [ \sqrt[i]{n}] = sum [\log_i n]

Source:

August 18, 2019
floor functionradicallogarithmSumalgebra

Problem Statement

Prove that for all natural numbers nn greater than 11 : [n]+[n3]+...+[nn]=[log2n]+[log3n]+...+[lognn][\sqrt{n}] + [\sqrt[3]{n}] +...+[ \sqrt[n]{n}] = [\log_2 n] + [\log_3 n] + ... + [\log_n n]
(VV Kisil)