Use the Comparison Test to show that ∞Σ(n=1) 1/[(n^3)+3] converges, and that ∞Σ(n=1) (6^n)/[(4^n)−1] diverges.?

Sol
1
Sow that Σ(n=1 to ∞)_1/(n^3+3) Converge
A=Σ(n=1 to ∞)_1/n^3
B=Σ(n=1 to ∞)_1/(n^3+3)
C=Σ(n=1 to ∞)_1/(n^3+3n^3)=Σ(n=1 to ∞)_1/(4n^3)
A>B>C
A is Converge
C is Converge
So
B is Converge
2
show that Σ(n=1 to ∞)_(6^n)/[(4^n)−1] Diverges
M=Σ(n=1 to ∞)_(6^n)/4^n=Σ(n=1 to ∞)_(3/2)^n
N=Σ(n=1 to ∞)_(6^n)/[(4^n)−1]
P=Σ(n=1 to ∞)_(6^n)/[(4^n)−(1/2)*4^n]=2*Σ(n=1 to ∞)_(3/2)^n
So
M<N<P
M is dIverge
P is Diverge
So
N is Diverge