1. Because we let a continuous variable instead of a discrete variable. And continuous variables can not be added with the power series. Thats why he used integral transformation

2. Because the sum will be a function of X (the variable)


3. A(n) will be same as before because it was just transformed into functional notation.

4. if a(n) = 1 and n = 0,1,2,3,4,5,6,......


then we get,

     1.x^0 + 1.x + 1.x^2 + 1.x^3 + 1.x^4 + 1.x^5 + ...
    = 1 + x + x^2 + x^3 + x^4 + x^5 + ...
    = 1/(1-x) [ we know that (1+x)^n = 1 + nx + [{n.(n-1)}/2!].x^2}.x^2] + .... + [{n(n-1)(n-2) ... (n-r+1)}/r!].x^r ]

5. if a(n)=1/n! and n = 0,1,2,3,4,5,6,......

then we get,

   1/0.x^0 + 1/1!.x^1 + 1/2!.x^2 + 1/3!.x^3 + 1/4!.x^4 + 1/5!.x^5 + ....
 = 1 + x + 1/2 x^2 + 1/6 x^3 + 1/24.x^4 + 1/120.x^5 + ... [ we know that (1+x) = 1 + nx + [{n.(n-1)}/2!].x^2 + .... + [{n(n-1)(n-2) ... (n-r+1)}/r!] x^r]

 = e^x

6. In summation equations, the value of n was discrete. Like n = 1,2,3,4..... the set of integers. But a continuous analog means it could represent any non negative number such as 1/2 , 3/2, Pi etc.

7. because no mathematician would like to have as the base of an exponential something like 'x'. Only convenient thing to
have is 'e' and the reason is its easy to integrate or differentiate.

8. Only for the values of |x|<1 the solution of x (1/1-X ) is true

9. Ln(x) will always prove a negative value. So for -s the value will be reverted back to positive.

10. Yes. I think learning in this way is pretty helpful. We can learn different things from this kind of way. It also improves our analytical power.