for-all-n-N-f-n-x-k-1-n-1-n-x-n-for-any-1-lt-x-lt-1-If-f-1-1-R-with-f-lim-n-f-n-of-1-1-0-1-2-f-x-dx-

Question Number 56708 by gunawan last updated on 22/Mar/19

for all n \in N, f_{n} (x) = \underset{k = 1}{\sum^{n}} {(- 1)}^{n} x^{n}

for any - 1 < x < 1 . If f : (- 1, 1) \to R

with f = \lim_{n \to \infty} f_{n} of (- 1, 1)

\int_{0}^{\frac{1}{2}} f (x) d x = \dots

Commented by Abdo msup. last updated on 22/Mar/19

w e h a v e f_{n} (x) = \sum_{k = 1}^{n} {(- x)}^{k} = \sum_{k = 0}^{n} {(- x)}^{k} - 1

= \frac{1 - {(- x)}^{n + 1}}{1 + x} - 1 = \frac{1 - {(- x)}^{n + 1} - 1 - x}{1 + x}

b u t ∣ x ∣< 1 \Rightarrow {l i m}_{n \to + \infty} f_{n} (x) = \frac{- x}{1 + x} \Rightarrow

f (x) = \frac{- x}{1 + x} \Rightarrow \int_{0}^{\frac{1}{2}} f (x) d x = - \int_{0}^{\frac{1}{2}} \frac{1 + x - 1}{1 + x} d x

= - \frac{1}{2} + \int_{0}^{\frac{1}{2}} \frac{d x}{1 + x} = - \frac{1}{2} + {[l n ∣ 1 + x ∣]}_{0}^{\frac{1}{2}}

= - \frac{1}{2} + l n (\frac{3}{2}) .

h e r e i h a v e u s e d t h a t l i m f_{n} = f .