What is the interval of convergence of `sum (x^n)/(n!)`?

For any `x in RR`, choose `N in ZZ` such that `N > abs(x)`

`abs(sum_(n=0)^oo x^n/(n!)) = abs(sum_(n=0)^(N-1) x^n/(n!) + sum_(n=N)^oo x^n/(n!)) <= sum_(n=0)^(N-1) abs(x)^n/(n!) + sum_(n=N)^oo abs(x)^n/(n!)`

`< sum_(n=0)^(N-1) abs(x)^n/(n!) + abs(x)^N/(N!) sum_(n=0)^oo abs(x)^n/(N^n)`

The first term `sum_(n=0)^(N-1) abs(x)^n/(n!)` is a finite sum so converges.

The second term `abs(x)^N/(N!) sum_(n=0)^oo abs(x)^n/(N^n)` is the sum of a geometric series with positive common ratio `abs(x)/N < 1`, so converges.

We have shown that for any `x in (-oo,oo)`, `sum_(n=0)^oo abs(x)^n/(n!)` is bounded, that is that `sum_(n=0)^oo x^n/(n!)` is absolutely convergent. Hence it is also convergent.

Take the absolute values and apply the ratio test

`lim_(n->oo)abs(a_(n+1))/(abs(a_n))=lim_(n->oo) (abs(x)^(n+1)/((n+1)!))/(abs(x)^(n)/((n)!))=lim_(n->oo) absx/((n+1)!)=0`

The limit is less than 1, independent of the value of x. It follows that the series converges for all x.
That is, the interval of convergence is `−∞ < x < +∞`.

Actually the sum is equal to the exponential function

`Σ x^n/(n!)=e^x`