海外大學試題2

The term of the above sequence is given by
a_n = n/(n+1)! = 1/n! - 1/(n+1)!

S_n
= Sum of (1/k! - 1/(k+1)!) (<- k = from 1 to n)
= Sum of (1/k!) - Sum of (1/(k+1)!)
= (1 + 1/2! + ... + 1/k!) - (1/2! + 1/3! + ... + 1/(k+1)!)
= 1 - 1/(k+1)!

Now
Infinite sum of (S_i x^i)
= Sum of ((1 - 1/(i+1)!)x^i)
= Sum of (x^i) - sum of (x^i/(i+1)!)

Note:
e^x = 1 + x + x^2/2! + x^3/3! + ...
e^x/x = 1/x + 1 + x/2! + x^2/3! + ...
sum of (x^i/(i+1)!) = x/2! + x^2/3! + ...

So the above sum is equal to
= Sum of x^i - e^x/x + 1/x + 1
= (1/x + 1 + x + x^2 + ...) - e^x/x
= 1/x(1-x) - e^x/x

The last equal sign converges only if |x| < 1 and e^x/x is defind only when x is nnon-zero. The given interval 0 < x < 1 satisfy the above condition.

Q.E.D.