determine if the series converges or diverges:
1/(ksqrt(k^2+1)) from k = 1 to infinity
as k→∞, √(k^2+1) ≈ k, so the terms approach 1/k^2 so it converges.
would you use the limit comparison test?
that would be good, since √(k^2+1) > √k^2 = k
im just confused on how to show my work with the limit comparison test. would my b_n = 1/sqrt(k^2) and then my a_n would be 1/(ksqrt(k^2+1))
and since b_n converges a_n will also converge?
yes, though you want b_n = 1/k^2
then a_n = 1/(k√(k^2+1)) < 1/(k√k^2) < 1/k^2 which converges.
To determine if the series converges or diverges, we can use the limit comparison test.
The limit comparison test compares the given series to a known series whose convergence is already known.
Let's consider the series you provided: 1/(k*sqrt(k^2+1)), from k=1 to infinity.
To apply the limit comparison test, we need to find a known series with positive terms that is easier to analyze and whose convergence is already known. In this case, we can choose the harmonic series: 1/k. The harmonic series is known to diverge, so if our series behaves similarly to it, the given series will also diverge.
Now, let's find the limit of the ratio of the two series as k approaches infinity:
lim(k->∞) (1/(k*sqrt(k^2+1))) / (1/k)
To simplify the expression, multiply both the numerator and denominator by k:
lim(k->∞) (k/(k*sqrt(k^2+1)))) / (1/k)
Cancel out the k terms:
lim(k->∞) (1/sqrt(k^2 + 1))
As k approaches infinity, the denominator becomes dominated by the k^2 term. Therefore, the limit becomes:
lim(k->∞) (1/sqrt(k^2)) = 1/k
Now, we can see that the limit is equivalent to the harmonic series, which means that the given series exhibits the same behavior as the harmonic series. Since the harmonic series diverges, this indicates that the original series also diverges.
Therefore, the given series 1/(k*sqrt(k^2+1)), from k=1 to infinity, diverges.