Cauchy sequences?

Question

Show that 1 - 1/3 + 1/5 - 1/7 + 1/9 - ...... is a Cauchy sequence. Show that the limit lies strictly between 2/3 and 13/15.

Answer 1

1 - 1/3 + 1/5 - 1/7 + ...

= Σ [1/(4n+1) - 1/(4n+3)] n = 0,1,2,...

= Σ [ ∫{0.. 1} x^(4n) dx - ∫{0.. 1} x^(4n+2) dx] n = 0,1,2,...

= Σ [ ∫ (1-x^2)x^4n dx ]

= ∫ (1 - x^2) Σ x^4n dx

= ∫ (1 - x^2) {1/(1 - x^4)} dx

= ∫{0.. 1} 1/(1 + x^2) dx
= Arctan(x) from 0 to 1...

= Arctan(1) - Arctan(0)
= 0.785398

since the series is convergent... then it is Cauchy. the series is in real numbers...


§