let-put-u-n-k-1-n-k-k-1-prove-that-u-n-n-1-1-2-study-the-convergence-of-n-1-1-u-n-

Question Number 32273 by abdo imad last updated on 22/Mar/18

l e t p u t u_{n} = \sum_{k = 1}^{n} k (k!)

1) p r o v e t h a t u_{n} = (n + 1)! - 1

2) s t u d y t h e c o n v e r g e n c e o f \sum_{n = 1}^{\infty} \frac{1}{u_{n}} .

Commented by abdo imad last updated on 25/Mar/18

1) l e t p r o v e t h i s r e l a t i o n b y r e c u r r e n c e

p (1) u_{1} = 2! - 1 = 1 \times (1!) p (1) i s t r u e l e t s u p o s e

u_{n} = (n + 1)! - 1 \Rightarrow u_{n + 1} = \sum_{k = 1}^{n + 1} k (k!) = \sum_{k = 1}^{n} k (k!) + (n + 1) (n + 1)!

= (n + 1)! - 1 + (n + 1) (n + 1)! = (n + 1)! (n + 1 + 1) - 1

= (n + 2) (n + 1)! - 1 = (n + 2)! - 1 t h e i m p l i c a t i o n

p (n) \Rightarrow p (n + 1) i s t r u e .

2) w e h a v e \sum_{n = 1}^{\infty} \frac{1}{u_{n}} = \sum_{n = 1}^{\infty} \frac{1}{(n + 1)! - 1} b u t f o r n \to \infty

\frac{1}{(n + 1)! - 1} \sim \frac{1}{(n + 1)!} a n d Σ \frac{1}{(n + 1)!} i s c o n v e r g e n t s o

Σ \frac{1}{u_{n}} i s c o v e r g e n t .

Answered by Tinkutara last updated on 23/Mar/18

u_{n} = \underset{k = 1}{\sum^{n}} (k + 1 - 1) (k!)

u_{n} = \underset{k = 1}{\sum^{n}} [(k + 1)! - k!]

I t t e l e s c o p e s a n d b e c o m e s

u_{n} = (n + 1)! - 1