use integral test to determine whether the series is convergent or divergent
infinity (sum symbol) n=1 1/(sqrt n)^5
To determine whether the series ∑(n=1 to infinity) 1/(sqrt(n))^5 is convergent or divergent, we can use the integral test. The integral test states that if a function f(x) is continuous, positive, and decreasing for x ≥ 1, and if a series ∑ a(n) converges if and only if the integral ∫ f(x) dx converges, then the series ∑ a(n) and the integral ∫ f(x) dx have the same convergence behavior.
In this case, we have the series ∑(n=1 to infinity) 1/(sqrt(n))^5. Let's start by checking if the conditions of the integral test are satisfied:
1. Continuity: The function f(x) = 1/(sqrt(x))^5 is continuous on x ≥ 1, as the square root function and constant multiplication preserve continuity.
2. Positivity: The function f(x) = 1/(sqrt(x))^5 is positive for x ≥ 1 since the denominator is always positive and nonzero.
3. Decreasing: To determine if the function f(x) = 1/(sqrt(x))^5 is decreasing for x ≥ 1, we can check its derivative. Taking the derivative of f(x), we have:
f'(x) = (-5/2)x^(-7/2).
Since f'(x) is negative for x ≥ 1, we can conclude that f(x) is decreasing for x ≥ 1.
Now that we have met all the conditions, we can proceed to calculate the integral:
∫ f(x) dx = ∫ 1/(sqrt(x))^5 dx
To evaluate this integral, we can simplify the expression inside the integral:
∫ x^(-5/2) dx
Using the power rule for integration, we have:
∫ x^(-5/2) dx = (x^(-5/2 + 1))/(-5/2 + 1) + C
= (-2/3) x^(-3/2) + C
Now, let's evaluate the integral from x = 1 to infinity:
∫[1,∞] f(x) dx = lim[b->∞] ∫[1,b] f(x) dx
= lim[b->∞] ((-2/3) x^(-3/2)) evaluated from x = 1 to x = b
= lim[b->∞] (-2/3) (b^(-3/2) - 1^(-3/2))
= lim[b->∞] (-2/3) (b^(-3/2) - 1)
= (-2/3) * 0 - (-2/3) * 1
= 2/3
Since the integral ∫ f(x) dx converges (equals 2/3), the series ∑(n=1 to infinity) 1/(sqrt(n))^5 also converges by the integral test.
Therefore, the series ∑(n=1 to infinity) 1/(sqrt(n))^5 is convergent.