Basic Comparison Test and Limit comparison test.

Question

What is the conclusion of the BCT and LCT when applied to: ∑ ∞, n=1  1/(³√(n²) + 9)

 

equation reads: summation of 1 goes to infinity.... one divided by cubed root of n squared plus nine.

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Answer 1

Use the BCT first to show that the related series ∑ ^∞ _n=1  1/(³√(n²) ) diverges:

 

³√(n²) = n^2/3 ≤ n, so n^-2/3 ≥ n^-1 for n=1,2,3,...

Therefore,

∑  1/(³√(n²) ) ≥ ∑  1/n

But ∑  1/n is the harmonic series, which we know diverges, so by the BCT test, ∑  1/(³√(n²) ) also diverges.

 

Now use the LCT to compare your series with ∑  1/(³√(n²) ):

 

lim_n→∞ ( (1/(³√(n²) + 9))/(1/(³√(n²) )) = lim ³√(n²) / (³√(n²) + 9) = 1

 

The limit is finite and positive, so by the LCT, since ∑  1/(³√(n²) ) diverges, ∑  1/(³√(n²)+9 ) also diverges.

Answer 2

The reasoning in Andre's answer is correct. You might review p-series, whose conclusion you could cite in your answer.

Anytime direct comparison works, limit comparison will also using the same function, but sometimes the limit comparison test will work when direct comparison seems too complicated, because you just need to find any other function whose corresponding series you know converges or diverges for the limit comparison test to yield an answer, AS LONG AS THE LIMIT OF THE RATIO IS NOT ZERO (emphasis added, because this precondition is often overlooked). With the aid of L'hopital's rule, the ratios of limits is often easy to find.