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
To determine whether the series converges or diverges, we can use the limit comparison test.
Step 1: Let's establish a comparison series that is easier to analyze. In this case, we can select the series 1/k, which converges (known as the harmonic series).
Step 2: Take the limit as k approaches infinity of the ratio of the terms of the two series:
lim(k->∞) ((√ln(k))/k) / (1/k)
Step 3: Simplify the expression:
lim(k->∞) (√ln(k))/(1/k)
Now, we can simplify this expression by multiplying both the numerator and the denominator by √(k):
lim(k->∞) (√ln(k))/(1/k) * (√k/√k)
After simplification, we get:
lim(k->∞) √(ln(k) * k)
Step 4: Evaluate the limit:
lim(k->∞) √(ln(k) * k)
As k approaches infinity, ln(k) and k both increase without bound. The product ln(k) * k will also increase without bound. And since the square root of an increasing function will also increase without bound, we can conclude that the limit of the ratio is infinity.
Step 5: Since the limit of the ratio is not finite (it is infinity), the given series diverges by the limit comparison test.
In summary, the series ∑((√ln(k))/k) diverges.