is this correct?
use the integral test to determine if this series is convergent or divergent:
the series from n=2 to infinity of 1/(n*square root of (ln(n)))
I said it was divergent because the integral went to infinity
To determine if the given series is convergent or divergent, we can indeed use the integral test. The integral test states that if f(x) is a continuous, positive, and decreasing function over the interval [N, ∞), where N is some positive integer, and if the function f(x) is integrable from N to ∞, then the series Σ f(n) from N to ∞ converges if and only if the improper integral ∫ f(x) dx from N to ∞ converges.
In this case, we have the series Σ (1/(n * √(ln(n)))) from n = 2 to ∞.
To verify the conditions of the integral test, let's check if the function f(x) = 1/(x * √(ln(x))) satisfies the required conditions:
1. f(x) is positive: Yes, since both x and √(ln(x)) are positive for x > 0, and dividing positive numbers will yield a positive value.
2. f(x) is decreasing: To determine the behavior of the function, let's take its derivative: f'(x) = (-1/(2x^2 * √(ln(x)))) - (1/(x^2 * 2√(ln(x)) * √(ln(x)))) = -((1 + 1)/(2x^2 * √(ln(x)))) = -1/(x^2 * √(ln(x))).
Since the derivative is negative for x > 0, it means f(x) is decreasing.
3. f(x) is integrable: To check if the integral, ∫ f(x) dx from 2 to ∞, exists, we need to analyze the behavior of the function as x approaches infinity. Taking the limit of f(x) as x approaches infinity will help us determine this:
lim(x->∞) (1/(x * √(ln(x)))) = 0,
Therefore, the integral ∫ f(x) dx from 2 to ∞ exists.
Now, to determine if the series converges or diverges, we evaluate the integral:
∫ (1/(x * √(ln(x)))) dx from 2 to ∞ = 2 * ∫ (1/(x * √(ln(x)))) dx from 1 to ∞ (substituting x = 2u)
Let's solve this integral using u-substitution:
Let u = ln(x), then du = 1/x dx.
When x = 1, u = ln(1) = 0, and when x = ∞, u = ln(∞) = ∞.
The integral becomes:
2 * ∫ (1/(x * √(ln(x)))) dx from 1 to ∞ = 2 * ∫ (1/(e^u * √(u))) du from 0 to ∞.
Considering the limits when integrating from 0 to ∞, the integral becomes:
2 * ∫ (1/(e^u * √(u))) du from 0 to ∞ = 2 * ∫ (1/(√(u) * e^u)) du from 0 to ∞.
Solving this integral is not straightforward; however, it can be shown that this integral does not converge. Therefore, the given series Σ (1/(n * √(ln(n)))) from n = 2 to ∞ diverges.
Hence, your conclusion that the series diverges is correct based on the integral test.