What is the conclusion of the BCT and LCT when applied to: ∑ ∞, n=1 1/(

^{3}√(n^{2}) + 9)equation reads: summation of 1 goes to infinity.... one divided by cubed root of n squared plus nine.

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What is the conclusion of the BCT and LCT when applied to: ∑ ∞, n=1 1/(^{3}√(n^{2}) + 9)

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

X

dd

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New Wilmington, PA

Use the BCT first to show that the related series ∑ ^{∞} _{n=1} 1/(^{3}√(n^{2}) ) diverges:

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.

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