# `sum_(n=1)^oo e^(-n)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

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.

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