Higher level maths

Consider: (1/101) + (1/102) + (1/103) + ...... + (1/10000)
Number of terms = 10000 - 101 + 1 = 9900
Each term ≥ (1/10000)

Hence, (1/101) + (1/102) + (1/103) + ...... + (1/10000) > [9900 x (1/10000)]
(1/101) + (1/102) + (1/103) + ...... + (1/10000) > (9900/10000)


(1/100) + (1/101) + (1/102) + (1/103) + ...... + (1/10000)
> (1/100) + (9900/10000)

But
= (1/100) + (9900/10000)
= (100/10000) + (9900/10000)
= 1

Hence,
(1/100) + (1/101) + (1/102) + (1/103) + ...... + (1/10000) > 1

∫[100 to 10000] (1/x) dx

= ln|x| [100 to 10000]

= ln(10000) - ln(100)

= 4.605 >1

2014-01-21 00:30:36 補充：
電腦程式計算的結果是 = 4.61022851840477

Private Sub Command1_Click()
valsum = 0
For cnt = 100 To 10000
valsum = valsum + (1 / cnt)
Next cnt
Text1.Text = valsum
End Sub