L-lim-i-1-i-1-i-i-1-L-

Question Number 4587 by FilupSmith last updated on 09/Feb/16

L = \lim_{i \to \infty} \frac{{(- 1)}^{i + 1} i}{i + 1}

L = ?

Commented by Yozzii last updated on 14/Feb/16

L d o e s n o t e x i s t . W h i l e ∣ L ∣= \lim_{i \to \infty} \frac{i}{i + 1}

∣ L ∣= \lim_{i \to \infty} \frac{1}{1 + \frac{1}{i}} = \frac{1}{1 + \frac{1}{\infty}} = \frac{1}{1 + 0} = 1

t h i s i n d i c a t e s t h a t f o r v e r y l a r g e i

t h e t e r m s o f {\frac{i}{1 + i}}_{i = 1}^{\infty} c o n v e r g e t o 1 .

H o w e v e r, {{(- 1)}^{i + 1} \frac{i}{1 + i}}_{i = 1}^{\infty} a l t e r n a t e s

p e r i o d i c a l l y b e t w e e n 1 a n d - 1 f o r

v e r y l a r g e i a c c o r d i n g t o t h e p a r i t y o f

i . S i n c e t h i s s e q u e n c e b e h a v e s i n a n

a l t e r n a t i n g m a n n e r c e a s e l e s s l y b e t w e e n

1 a n d - 1 (w i t h p e r i o d 2) a s i \to \infty,

w h e r e i f L e x i s t s i t t a k e s o n e v a l u e,

t h e l i m i t L o f t h e s e q u e n c e {\frac{{(- 1)}^{i + 1} i}{i + 1}}_{i = 1}^{\infty} d o e s n o t

e x i s t .

Facebook Tweet Pin