e-x-2-dx-as-an-infinite-series-Hence-investigate-its-converge-

Question Number 56060 by necx1 last updated on 09/Mar/19

\int e^{- x^{2}} d x a s a n i n f i n i t e s e r i e s . H e n c e

i n v e s t i g a t e i t s c o n v e r g e .

Commented by maxmathsup by imad last updated on 09/Mar/19

w e h a v e e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!} \Rightarrow \int e^{- x^{2}} d x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} \int x^{2 n} d x

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

l e t u_{n} (x) = \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1) n!} \Rightarrow f o r x \neq 0

∣ \frac{u_{n + 1} (x)}{u_{n (x)}} ∣ =∣ \frac{∣ x ∣^{2 n + 3}}{(2 n + 3)! (n + 1)!} \frac{(2 n + 1) n!}{x^{2 n + 1}} ∣ = \frac{2 n + 1}{(2 n + 3) (n + 1)} ∣ x ∣^{2} \to 0 (n \to + \infty) \Rightarrow

\Rightarrow R = + \infty .

Commented by necx1 last updated on 09/Mar/19

T h a n k s f o r t h e h e l p t h o I r e a l l y d o n o t

u n d e r s t a n d . P l e a s e s h e d m o r e l i g h t

Commented by maxmathsup by imad last updated on 09/Mar/19

s i r t a k e a l o o k a t c k n v e r g e n c e o f s e r i e s u b j e c t \dots