find-the-radius-of-convergence-for-the-serie-n-0-e-n-z-n-z-from-C-

Question Number 26176 by abdo imad last updated on 21/Dec/17

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

s e r i e \sum_{n = 0}^{\propto} e^{- \sqrt{n}} z^{n} \dots z f r o m C .

Commented by abdo imad last updated on 23/Dec/17

w e p u t u_{n (z)} = e^{- \sqrt{n}} z^{n} a n d f o r z n o t o

/ \frac{u_{n + 1 (z)}}{u_{n (z)}} / = / \frac{e^{- \sqrt{n + 1}} z^{n + 1}}{e^{- \sqrt{n}} z^{n}} / = e^{\sqrt{n} - \sqrt{n + 1}} / z /

b u t l i m_{n - >\propto} e^{\sqrt{n} - \sqrt{n + 1}} = e^{\frac{- 1}{\sqrt{n} + \sqrt{n + 1}}} = 1 f i n a l l y

t h e s e r i e c o n v e r g e \Leftrightarrow / z / < 1

i f z = - 1 Σ e^{- \sqrt{n_{}}} {(- 1)}^{n} i s c o n v e r g e n t b e c a u s e i t s a a l t e r n a t i n g s e r i e

f o r z = 1 t h e s e r i e Σ e^{- \sqrt{n}} i s c o n v e r g e n t b e c a u s e t h e i n t e g r a l

\int_{0}^{\infty} e^{- \sqrt{t}} d t i s c o n v e r g e n t (t h e f u n c t i o n ψ (t) = e^{- \sqrt{t}} i s d e c r e a s i n g

i n [0 . \propto [\Rightarrow i n t e r v a l o f c o n v e r g e n c e i s D_{c} = [- 1.1]