Use the Integral Test to determine whether the series ∑∞,n=1 1/(n^2+n^3) is convergent. (First make sure to check that the conditions of the Integral Test are satisfied.)
To use the Integral Test to determine the convergence of a series, we need to check two conditions:
1. The terms of the series are positive: In this case, the terms of the series are 1/(n^2 + n^3) which is positive for all n > 0.
2. The function f(x) = 1/(x^2 + x^3) is continuous, positive, and decreasing on the interval [1, ∞): To check this, we can take the derivative of f(x) and see if it's negative.
Let's calculate the derivative of f(x):
f'(x) = d/dx (1/(x^2 + x^3))
= -1/(x^2 + x^3)^2 * (2x + 3x^2)
Now we need to find the critical points of f(x) by finding where f'(x) = 0:
-1/(x^2 + x^3)^2 * (2x + 3x^2) = 0
From this equation, we can see that the only possible critical point is x = 0. However, since x > 1, the function is always positive and decreasing on the interval [1, ∞).
Now, we can apply the Integral Test:
∫[1, ∞] 1/(x^2 + x^3) dx
To evaluate this integral, we can use partial fraction decomposition. The denominator x^2 + x^3 factors to x^2(1 + x), so we can write:
1/(x^2 + x^3) = A/x + B/(1 + x)
Multiplying both sides by x(x+1), we get:
1 = A(1 + x) + Bx
Simplifying further:
1 = (A + B) + (Ax + Bx)
1 = A + B + Ax + Bx
Comparing coefficients:
A + B = 0 (coefficient of constant term on the right is 0)
A + B = 1 (coefficient of x term on the right is 1)
From these equations, we find A = -1 and B = 1.
Thus, the original integral becomes:
∫[1, ∞] (-1/x + 1/(1 + x)) dx
Taking the antiderivative:
-∫[1, ∞] 1/x dx + ∫[1, ∞] 1/(1 + x) dx
= -ln|x| + ln|1 + x| |[1, ∞]
Evaluating the limits:
-ln|∞| + ln|1 + ∞| - (-ln|1| + ln|1 + 1|)
= -ln|1 + ∞| + ln|2| - (-ln|1| + ln|2|)
= -ln(∞) + ln(2) + ln(2)
= -∞ + ln(2) + ln(2)
= -∞
Since the integral is divergent (equals -∞), the series ∑∞,n=1 1/(n^2+n^3) is also divergent based on the Integral Test.