Student Question

# 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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