Use the BCT first to show that the related series ∑ ^{∞} _{n=1} 1/(^{3}√(n^{2}) ) diverges:
^{3}√(n^{2}) = n^{2/3} ≤ n, so n^{2/3} ≥ n^{1} for n=1,2,3,...
Therefore,
∑ 1/(^{3}√(n^{2}) ) ≥ ∑ 1/n
But ∑ 1/n is the harmonic series, which we know diverges, so by the BCT test, ∑ 1/(^{3}√(n^{2}) ) also diverges.
Now use the LCT to compare your series with ∑ 1/(^{3}√(n^{2}) ):
lim_{n→∞} ( (1/(^{3}√(n^{2}) + 9))/(1/(^{3}√(n^{2}) )) = lim
^{3}√(n^{2}) / (^{3}√(n^{2}) + 9) = 1
The limit is finite and positive, so by the LCT, since ∑ 1/(^{3}√(n^{2}) ) diverges, ∑ 1/(^{3}√(n^{2})+9 ) also diverges.
10/22/2013

Andre W.