Convergence Test for Σ 1/(n^2 logn)

In order to test the convergence of the series Σ 1/(n^2 logn), we can use the Integral Test. The Integral Test states that if f(x) is a positive, continuous, and decreasing function for x ≥ 1, then the series Σ f(n) converges if and only if the integral ∫ f(x) dx from 1 to ∞ converges.

Step 1: Check the function f(x)
In this case, our function f(x) = 1/(x^2 logx) is positive, continuous, and decreasing for x ≥ 2.

Step 2: Evaluate the integral
Now, we need to find the integral of f(x) from 2 to ∞:
∫ 1/(x^2 logx) dx = ∫ 1/u du (letting u = logx)
= ∫ 1/u du
= ln|u| + C
= ln|logx| + C

Step 3: Determine the convergence
To determine the convergence of the integral, we need to check if the integral ln|logx| converges as x approaches infinity. Since ln|logx| grows very slowly, the integral ln|logx| actually diverges as x approaches infinity.

Conclusion
Since the integral of the function 1/(n^2 logn) diverges, by the Integral Test, the series Σ 1/(n^2 logn) also diverges. Therefore, the series does not converge.