# How do you use the integral test to determine if #Sigma e^(-n/2)# from #[1,oo)# is convergent or divergent?

The sum

We must calculate the improper integral

First we compute the integral

Perform the substitution

Therefore,

Now, compute the boundaries

Therefore,

The integral is convergent.

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To determine if the series ( \sum_{n=1}^{\infty} e^{-\frac{n}{2}} ) from ( n = 1 ) to infinity is convergent or divergent using the integral test, follow these steps:

- Consider the function ( f(x) = e^{-\frac{x}{2}} ).
- Verify that ( f(x) ) is continuous, positive, and decreasing for ( x \geq 1 ).
- Apply the integral test by integrating ( f(x) ) from 1 to infinity:

[ \int_{1}^{\infty} e^{-\frac{x}{2}} , dx ]

- Evaluate the integral.
- If the integral converges, then the series converges. If the integral diverges, then the series diverges.

To evaluate the integral:

[ \int_{1}^{\infty} e^{-\frac{x}{2}} , dx ]

Use substitution: Let ( u = -\frac{x}{2} ), then ( du = -\frac{1}{2} , dx ). This leads to ( dx = -2 , du ).

So, the integral becomes:

[ \int e^u \cdot (-2 , du) ]

[ = -2 \int e^u , du ]

[ = -2e^u + C ]

[ = -2e^{-\frac{x}{2}} + C ]

Evaluate the integral from 1 to infinity:

[ -2e^{-\frac{x}{2}} \Bigg|_{1}^{\infty} ]

[ = -2e^{-\frac{\infty}{2}} - (-2e^{-\frac{1}{2}}) ]

[ = -2 \cdot 0 - (-2e^{-\frac{1}{2}}) ]

[ = 2e^{-\frac{1}{2}} ]

Since ( 2e^{-\frac{1}{2}} ) is a finite value, the integral converges, and by the integral test, the series ( \sum_{n=1}^{\infty} e^{-\frac{n}{2}} ) is convergent.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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