p. 643

33.

a) Draw illustrations to show \(TraditionalForm
      \`ln(n + 1)\  = 
        \(\(∫\_1\%\(n + 1\)\(1\/x\) dx ≤ 
                  1 + 1\/2 +  ... \) + 1\/n ≤ 
            1 + ∫\_1\%n\( 1\/x\) dx = 
          1 + ln(n)\)\)

\(p1 = 
      Plot[1/Floor[x], {x, 1, 10}, \ PlotRange -> {{0, 10}, {0, 1}}, 
	
        PlotStyle -> RGBColor[0, 1, 0]]\)

\(p2 = 
      Plot[1/Floor[x + 1], {x, 1, 10}, \ PlotRange -> {{0, 10}, {0, 1}}, 

        		PlotStyle -> RGBColor[0, 0, 1]]\)

p  =  Plot[1/x,  {x, 1, 10},  PlotRange -> {{0, 10}, {0, 1}}]

The area under the green curve equals 1 + (1/2) +  ...  + (1/n)which is clearly greater than the area under the black curve given by \(TraditionalForm
      \`∫\_1\%\(n + 1\)\(1\/x\) dx\).  Therefore, \(TraditionalForm
      \`\(∫\_1\%\(n + 1\)\(1\/x\) dx ≤ 
            1 + 1\/2 + 1\/3 +  ... \) + 1\/n\).

Show[p1, p]

The area under the blue curve is (1/2) + (1/3) +  ...  + (1/n) which is clearly less than the area under the black curve gievn by (∫_1^n) (1/x) dx. Therefore, \(TraditionalForm
      \`\(1 + 1\/2 + 1\/3 +  ... \) + 1\/n\  ≤ 
        1 + ∫\_1\%n\( 1\/x\) dx\).

Show[p2, p]

b) Find how large the partial sum (Subscript[s, n]) = 1 + (1/2) +  ...  + (1/n) is when n equals the number of seconds since the beginning of the universe.

From part a) \(TraditionalForm
      \`ln(n + 1) ≤ s\_n ≤ 1\  + \ ln(n)\). The number of seconds in a year is

yr  =  365*24*60*60

Therefore, the number of seconds in 13 billion years is approximately

bign  =  13×(10^9)* yr

The bounds on (Subscript[s, n])are:

Log[bign + 1] // N1 + Log[bign] // N

So (Subscript[s, n])is only between 40.55 and 41.56. Still a long way to ∞!

Actually, Mathematica has no trouble computing the value from the sum (notice the numerator has a decimal point to ensure the calculation is done numerically and not symbolically):

s[n_]  :=  (∑_(k = 1)^n) (1.0/k)

s[bign]