determine whether the series converges of diverges
the sum from k=2 to infinity of
(the square root of (ln(k)))/k
I said that because you can't integrate the series (goes to infinity) it diverges
is this true?
Yes, but did we not do this before using integral?
To determine whether the series converges or diverges, we can use the integral test. The integral test states that if a series ∑(aₙ) converges when the corresponding improper integral ∫(a(x))dx converges, and if the integral diverges, then the series also diverges.
In this case, we have the series ∑[(√ln(k))/k] from k=2 to infinity.
To apply the integral test, we need to check the convergence of the integral ∫[(√ln(x))/x]dx from x=2 to infinity.
To integrate this, we can make a substitution: let u = ln(x). Then, du/dx = 1/x, and dx = du/(1/x) = x du.
Substituting back into the integral, we have:
∫[(√ln(x))/x]dx = ∫√u du.
Integrating this yields: (2/3)u^(3/2) = (2/3)ln(x)^(3/2).
To evaluate this from x=2 to infinity, we find the limit as x approaches infinity of [(2/3)ln(x)^(3/2)], which is infinity.
Since the corresponding integral diverges, by the integral test, the series ∑[(√ln(k))/k] also diverges.
Therefore, your statement that the series diverges is correct.