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

Question

Ezra Adams · Accepted Answer

The sum #sum_1 ^ooe^(-2/n)# is convergent

We must calculate the improper integral #int_1^ooe^(-x/2)dx# First we compute the integral #I=inte^(-x/2)dx# Perform the substitution #u=-x/2#, #=>#, #du=-(dx)/2# Therefore, #I=int-2e^udu=-2e^u=-2e^(-x/2)+C# Now, compute the boundaries #lim_(x->1^+)-2e^(-x/2)=-2e^(-1/2)=-2/sqrte# #lim_(x->+oo)-2e^(-x/2)=0# Therefore, #int_1^ooe^(-x/2)dx=2/sqrte# The integral is convergent.

William Abdalla · Answer

undefined 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:

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

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

4. Evaluate the integral.
5. 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.