Determine whether the series is divergent or convergent, and if convergent, then absolutely or conditionally.

Question

Determine whether the series is divergent or convergent, and if convergent, then absolutely or conditionally. ∑from ∞ to n=2 (2lnn)/(n^2). Please help ASAP

Answer 1

Hi, Soyeb! For this series, you can use the Direct Comparision Test (DCT). Let's set the original sequence [2ln(n)]/(n²) as equal to a_n.

First, we need to find the parent sequence of this series, which we can label as c_n. The parent sequence will act similarly to the original sequence, but be easier to show the convergence/divergence of. The original sequence is [2ln(n)]/(n²). Notice that the denominator has a p-series (1/n^p). This is a series we can easily recognize and show the convergence/divergence of. Thus, we should set the parent sequence to be c_n = 1/n².

Now, we need to show the convergence/divergence of ∑c_n using either GST or PST. Like we learned earlier, this is a p-series, so we should use PST (p-series test). Our series is ∑1/n² , so p = 2. Since p is greater than 1, ∑c_n converges by PST.

Since the parent series converges, we need to find b_n, which is a sequence that is a multiple of the parent sequence, and satisfies the condition a_n ≤ b_n. So b_n needs to be bigger than the original sequence.

Let's say b_n = 1/(n)^3/2. This sequence is similar to the parent sequence, but since the denominator will generally be smaller, the sequence as a whole will be bigger. We should still test our condition:

a_n ≤ b_n --> 2ln(n)/n² ≤ 1/(n)^3/2.

This simplifes to 2ln(n) ≤ n^1/2.

If you test this by plugging in various values of n, you will find that this inequality holds true for n ≥ 10. For DCT, all that matters is that a_n ≤ b_n for all n greater than some integer N. Our inequality upholds for all n greater than N = 9.

Finally, we need to show the convergence/divergence of ∑b_n using either GST or PST. b_n = 1/(n)^3/2 is a p-series, so let's use PST. p = 3/2, which is greater than 1, so ∑b_n converges by PST.

Since ∑b_n converges and a_n ≤ b_n for all n greater than N = 9, then ∑a_n converges according to DCT.

Now that we know this series is convergent, to determine if it is convergent absolutley, we have to find the convergence of ∑|a_n|.

|a_n| = |2ln(n)]/(n²)|, which just equals 2ln(n)/n², because n is always positive. As you can see, a_n is the same as |a_n|. So ∑|a_n| will equal ∑a_n, meaning ∑|a_n| is also convergent.

Since ∑|a_n| and ∑a_n both converge, then ∑a_n converges absolutely.

Answer 2

Converge Absolutely

LCT: an = 1/n^2, bn = (2lnn)/(n^2)

lim (n->inf) an / bn = lim (n->inf) 1/n^2 / (2lnn)/(n^2) = lim (n->inf) 1 / (2lnn) = 0

an = 1/n^2 => conv by p series p>1, by LCT bn also conv.