find 1 /2! + 2 /3! + 3 /4! + ............ to infinity?

The typical terms of the sum are:

2/3!, 3/4!, 4/5!,........... In other words:

a(n) = (n-1)/n!

(where we are running the sum from n=2 to infinity).

But notice that

a(n) = n/n! - 1/n!

= 1/(n-1)! - 1/n!

Therefore, Sum(n=2, infinity) [a(n)]

= Sum(n=2, infinity) [(1/(n-1)! - 1/n!]

= (1/1! - 1/2!) + (1/2! - 1/3!) + (1/3! - 1/4!) +...

= 1/1! + (-1/2! + 1/2!) + (-1/3! + 1/3!) + (-1/4! +...)

= 1/1! + 0 + 0 + 0 + 0...

= 1/1! = 1

Doing this iteratively, the first term is:

T1 = 1/2

T1 = 1 - 1/2

T1 = 1 - 1/2!

The second term is:

T2 = 1/2 + 2/3!

T2 = 1/2 + 2/6

T2 = 1/2 + 1/3

T2 = 5/6

T2 = 1 - 1/6

T2 = 1 - 1/3!

The third term is:

T3 = T2 + 3/4!

T3 = T2 + 3/24

T3 = T2 + 1/8

T3 = 5/6 + 1/8

T3 = 20/24 + 3/24

T3 = 23/24

T3 = 1 - 1/24

T3 = 1 - 1/4!

As you can see, for any term N, its value is:

TN = 1 - 1/(N+1)!

For N at infinity, this is:

Tinf = 1 - 1/(inf+1)!

Tinf = 1 - 1/inf

Tinf = 1

1/2! + 2/3! + 3/4! +4/5! + --------

Note that 1 /2! = (2! -1) / 2! . Now, 1 /2! + 2 /3! + 3 /4! + ... + (n-1) / n! = (n! -1) / n! by induction because (n! -1) / n! + n / (n+1)! = ((n+1)! - (n+1)+n) / (n+1)!= ((n+1)!-1) / (n+1)! as desired. Hence 1 /2! + 2 /3! + 3 /4! + ............ = limit (n!-1) / n! =1.

1 /2! + 2 /3! + 3 /4! + ............ to infinity = 1 (definition)

26/12 ! = no answer. Since it would take an infinite amount of time to get to infinity. Infinity is much longer than any ones life span and any computer used to try to arrive at the answer would need an infinite amount of voltage,power,energy etc. To arrive at the answer. By then the Universe will have contracted to infinite mass or the primeval atom.

the ans is infinity