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Basic Comparison Test and Limit comparison test.

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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1 Answer

Use the BCT first to show that the related series ∑  n=1  1/(3√(n2) ) diverges:
 
3√(n2) = n2/3 ≤ 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) ):
 
limn→∞ ( (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.