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**.