Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 5

Recall the Integral test is applicable if f is positive and decreasing function on interval [k,oo) where a_n = f(x) .
If int_k^oo f(x) dx is convergent then the series sum_(n=k)^oo a_n is also convergent.
If int_k^oo f(x) dx is divergent then the series sum_(n=k)^oo a_n is also divergent.
For the series sum_(n=1)^oo e^(-n) , we have a_n=e^(-n) then we may let the function:
f(x) =e^(-x) with a graph attached below.
As shown on the graph, f(x) is positive and decreasing on the interval [1,oo) . This confirms that we may apply the Integral test to determine the converge or divergence of a series as:
int_1^oo e^(-x)dx =lim_(t-gtoo)int_1^t e^(-x)dx
To determine the indefinite integral of int_1^t e^(-x) dx , we may apply u-substitution by letting: u =-x then du = -dx or -1du =dx.
The integral becomes:
int e^(-x) dx =int e^u *( -1 du)
= - int e^u du
Apply integration formula for exponential function: int e^u du = e^u+C
- int e^u du =- e^u
Plug-in u =-x on - e^u , we get:
int_1^t 1/2^x dx= -e^(-x)|_1^t
= - 1/e^x|_1^t
Applying definite integral formula: F(x)|_a^b = F(b)-F(a).
- 1/e^x|_1^t = [- 1/e^t] - [- 1/e^1]
=- 1/e^t+ 1/e
Apply int_1^t e^(-x) dx=- 1/e^t+ 1/e , we get:
lim_(t-gtoo)int_1^t e^(-x) dx=lim_(t-gtoo)[- 1/e^t+ 1/e]
=lim_(t-gtoo)- 1/e^t+lim_(t-gtoo) 1/e
= 0 +1/e
=1/e or 0.368 (approximated value)
The lim_(t-gtoo)int_1^t e^-x dx=1/e implies the integral converges.
Conclusion:
The integral int_1^ooe^(-x) dx is convergent therefore the series sum_(n=1)^oo e^(-n) must also be convergent.