Series Test. Find convergence or divergence.

Find out if the sum of sqrt(n)/(n^3+1) as n goes from 1 to infinity is divergent or convergent.

We need to find a function, that we already know as convergent or divergent, that we can compare to. By looking at the leading n values, we can extract a function that is 1/n^(something) that is easily solved.

Rewrite the expression with only the leading/dominant values of n. We will get (n^(1/2))/(n^3).
Using the laws of exponents, this simplifies to 1/n^(5/2).
This series is convergent by the P-series test, which states that a series of 1/n^P is convergent when P > 1. Our P in this example is 5/2.
We can then use the Limit Comparison Test (LCT) to compare the original series with our P-series. The LCT states that if both series have positive terms and that dividing them gives us a finite number (C), and that finite number (C) is greater than zero then either both series diverge or converge.
We take the limit of our original series divided by our P-series comparison sum. We can see that when we simplify this down we get C=1. By the LCT these functions are either both convergent or divergent. Because our p-series comparison is convergent, our function is convergent.