study-the-derivability-of-f-x-n-0-1-n-nx-1-

Question Number 39378 by maxmathsup by imad last updated on 05/Jul/18

s t u d y t h e d e r i v a b i l i t y o f

f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n x + 1}

Commented by math khazana by abdo last updated on 10/Jul/18

t h e f u n c t i o n s f_{n} (x) = \frac{{(- 1)}^{n}}{n x + 1} a r e d e r i v a b l e s

i f w e c o n s i d e r ? f o r x f i x e d φ (t) = \frac{1}{x t + 1}

φ i s d e c r e a s i n g s o f (x) i s a a l t e r n a t e s e r i e

c o n v e r g e n t a l s o t h e s e r i e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(n x + 1)}^{2}} i s

u n i f . c o n v e r g e n t b e c a u s e ∣ \frac{{(- 1)}^{n + 1}}{{(n x + 1)}^{2}} ∣⩽ \frac{1}{n^{2} x^{2}}

s o f i s d e r i v a b l e a n d

f^{^{'}} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(n x + 1)}^{2}} .