How do you show whether #sum_(n=2)^oo 1/ln^3(n)# converges or diverges?

Answer 1

#sum_(n=2)^oo1/ln^3(n)# diverges by the Cauchy condensation test

The Cauchy condensation test states that if #f(n)# is nonnegative and nonincreasing real sequence, then #sum_(n=1)^oof(n)# converges if and only if the sum #sum_(n=0)^oo2^nf(2^n)# converges as well.
Let #f(n) = {(1/ln^3(n) if n>1),(f(n+1) if n=1):}#
As adding a finite value to a series does not change whether it converges or diverges, we can add #1/ln^3(2)# to the given series without changing the result. Doing so, we can apply the Cauchy condensation test:
#1/ln^3(2)+sum_(n=2)^oo1/ln^3(n) = sum_(n=1)^oof(n)# converges if and only if #sum_(n=0)^oo2^nf(2^n) = sum_(n=1)^oo2^n/ln^3(2^n)# converges.

Now, looking at the condensed sum, we have

#sum_(n=1)^oo2^n/ln^3(2^n) = sum_(n=1)^oo2^n/(nln(2))^3=sum_(n=1)^ooln^(-3)(2)2^n/n^3#
which diverges by the divergence test, as #lim_(n->oo)|2^n/n^3|=oo#.
As #sum_(n=0)^oo2^nf(2^n)# diverges, so does #sum_(n=1)^oof(n)=1/ln^3(2)+sum_(n=2)^oo1/ln^3(n)#, and thus so does the given series.
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Answer 2
My first intuition is to show that #1/ln^3(n)>1/n# to claim that #sum_(n=2)^oo1/ln^3(n)# diverges through direct comparison, since #sum_(n=2)^oo1/n# diverges as the harmonic series.
Notice that #1/ln^3(n)=1/n(n/ln^3(n))#. From this we see that #1/ln^3(n)>1/n# if #n/ln^3(n)>1#.
To see if this is true as these series extend infinitely, we can take the infinite limit of #n/ln^3(n)#.
#lim_(nrarroo)n/ln^3(n)#
This is indeterminate in the form #oo/oo#, so we can apply L'Hospital's rule and tale the derivative of the numerator and denominator separately.
#=lim_(nrarroo)1/(3ln^2(n)1/n)=lim_(nrarroo)n/(3ln^2(n))#

Reapplying L'Hospitals (we can see a pattern forming):

#=lim_(nrarroo)1/(6ln(n)1/n)=lim_(nrarroo)n/(6ln(n))#

L'Hospital's once more:

#=lim_(nrarroo)1/(6 1/n)=lim_(nrarroo)n/6=oo#
Thus we've seen that #n/ln^3(n)# will be far greater than #1#, and that we can state that #1/ln^3(n)=1/n(n/ln^3(n))>1/n# for sufficiently large values of #n#.
Through direct comparison, #sum_(n=2)^oo1/ln^3(n)# diverges.
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Answer 3

To determine whether the series (\sum_{n=2}^{\infty} \frac{1}{\ln^3(n)}) converges or diverges, we can use the integral test.

Let (f(x) = \frac{1}{\ln^3(x)}).

We integrate (f(x)) from (x = 2) to (x = \infty), which gives us: [\int_{2}^{\infty} \frac{1}{\ln^3(x)} dx]

To evaluate this improper integral, we can use substitution. Let (u = \ln(x)), then (du = \frac{1}{x} dx). Substituting these values, the integral becomes: [\int \frac{1}{u^3} du]

Integrating, we get: [\lim_{b \to \infty} \left[-\frac{1}{2u^2}\right]_{\ln(2)}^{\ln(b)}]

Evaluating the limit, we have: [\lim_{b \to \infty} \left[-\frac{1}{2\ln^2(b)} + \frac{1}{2\ln^2(2)}\right]]

Since (\lim_{b \to \infty} \frac{1}{2\ln^2(b)} = 0), the integral converges if and only if (\frac{1}{2\ln^2(2)}), a constant, converges. Since the constant is finite, the integral converges, and by the integral test, the series (\sum_{n=2}^{\infty} \frac{1}{\ln^3(n)}) converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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