#E(n) = (sum_(i=1)^n 1/i) / n = (H(n))/n#
where #H(n)# is the #n#th Harmonic number.
As #n -> oo#, #H(n) - ln(n) -> gamma#, where #gamma# is the Euler-Mascheroni constant. #gamma ~~ 0.5772#
So #H(n) ~~ ln(n) + gamma#
Actually a better approximation is #H(n) ~~ ln(n+1) + gamma#
Hence #E(n) ~~ (ln(n+1) + gamma)/n#
Using a spreadsheet, I found:
#E(5) = (sum_(i=1)^5 1/i) / 5 = 0.45dot(6)#
#E(100) = (sum_(i=1)^100 1/i) / 100 ~~ 0.05187#
Note that the approximation #E(n) ~~ (ln(n+1)+gamma)/n# gives us corresponding values: #0.4378# and #0.05192#
Background
The series:
#sum_(i=1)^oo 1/i = 1 + 1/2 + 1/3 + 1/4 +...#
is called the Harmonic Series.
It is well known that this does not converge, but how quickly does it diverge?
The finite sum #H(n) = sum_(i=1)^n 1/i# is called the #n#th Harmonic Number.
#H(n) = sum_(i=1)^n 1/i = int_1^(n+1) 1/floor(x) dx#
Compare this with the integral:
#int_1^(n+1) 1/x dx = ln(n+1)#
The smooth graph of #1/x# touches each of the steps of #1/floor(x)#.
The total area between these two curves between #1# and #oo# is the Euler-Mascheroni constant #gamma ~~ 0.5772156649#
So we can write:
#gamma = int_1^oo (1/floor(x) - 1/x) dx#
