How do i know which one is absolutely convergent, conditionally convergent or divergent?
sum_(n=2)^infinity ((-1)^n (n^3-6 n))/(8 n^3+1)
Answer:
Think of absolute convergence as the convergence of the series obtain by taking the absolute value of sum_(n=2)^infinity ((-1)^n (n^3-6n)/(8n^3+1)). What happens when we take the absolute value of each term of this series? Well for sure we know that the (-1)^n must go away since the absolute value of (-1)^n is always just 1. Next we note that (n^3-6n)/(8n^3+1) is always positive for all n in [2,infinity). Therefore absolut_value((-1^n)*(n^3-6n)/(8n^3+1))=(n^3-6n)/(8n^3+1). This means every term in the series is given by (n^3-6n)/(8n^3+1). If sum_n=1^infinity((n^3-6n)/(8n^3+1)) is a convergent series, we say the original series is absolutely convergent.
It is a property of absolute convergence that if the original series is absolutely convergent, then the series is also conditionally convergent. Therefore it only makes sense to test if the series is conditionally convergent when we know that the original series is not absolutely convergent. (Note: absolute convergence always implies conditional convergence, but conditional convergence does not always imply absolute convergence). So what does conditionally convergent mean?
Conditional convergence means that under the special “condition” of considering the series as either the alternating series or by multiplying by some constant we can obtain a series that is convergent (whereas the the original series is not convergent).
Divergence occurs when the limit of the series is infinity or negative infinity. This always occurs when the limit of the sequence (not series but SEQUENCE) has a limit going to infinity as n aproaches infinity, but this is not always necessary for divergence (as an example, consider that the infinite series from 1/n is divergent but lim(1/n)=0 as n approaches infinity).
