Question

Determine whether the series Σ_{n=2}[infinity] {1}{n√ {ln n}} is convergent or divergent using integral test showing that the scrics is positive and decreasing.

Answer 1

The series Σ from n=2 to infinity of 1/(n√{ln n}) is convergent if the improper integral of the corresponding function from 2 to infinity converges. This is determined by showing that the function is positive, continuous, and decreasing, and then evaluating the integral.

Step-by-step explanation:

The question asks whether the series Σ from n=2 to infinity of 1/(n√{ln n}) is convergent or divergent using the integral test. The integral test requires that the function be positive, continuous, and decreasing for all x >= some number N (in this case, N=2). To apply the integral test, we consider the improper integral of the function 1/(x√{ln x}) from 2 to infinity. We will show that this improper integral is convergent, which implies that the original series is also convergent.

Let's set up the integral:

1. ∫ from 2 to infinity of (1/x√{ln x}) dx.
2. First, we can see that the function is positive for x >= 2 because ln x is positive for x > 1, and we're only considering x >= 2.
3. Next, we should show that the function is decreasing. To do so, we can take the derivative and show that it's negative for x >= 2. The derivative of 1/(x√{ln x}) is negative because the numerator (ln x) increases as x increases, but at a slower rate than x√{ln x} does, so the overall function's rate of increase is slowing down, implying that the function itself is decreasing.
4. Assuming we have shown the function to be positive and decreasing, we would then evaluate the integral. If the integral from 2 to infinity is finite, then by the integral test, the series Σ from n=2 to infinity of 1/(n√{ln n}) is convergent.

Assuming we evaluated the integral and found it converges, we conclude that the series is indeed convergent.