How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n))/(sqrt(n+4))# from #[1,oo)#?

Question

Christopher Agresta · Accepted Answer

The series:

#sum_(n=1)^oo (-1)^n/sqrt(n+4) #

is convergent but not absolutely convergent.

This is an alternating series in the form: #sum_(n=1)^oo (-1)^na_n#, thus based on Leibniz's theorem, it is convergent if: #(1) lim_(n->oo) a_n = 0# #(2) a_n >= a_(n+1)# In our case: #lim_(n->oo) 1/sqrt(n+4) = 0# and #1/sqrt(n+4) > 1/sqrt(n+5)# so both conditions are satisfied and the series is convergent. Consider now the series of absolute values: #(3) sum_(n=1)^oo 1/sqrt(n+4)# Using the limit comparison test we can see that as: #lim_(n->oo) sqrt(n)/sqrt(n+4) = 1# the series #(3)# has the same character as the series: #sum_(n=1)^oo 1/sqrt(n) = sum_(n=1)^oo 1/n^(1/2)# which is divergent based on the #p#-series test. The series #(3)# is then also divergent, so the series: #sum_(n=1)^oo (-1)^n/sqrt(n+4) # is convergent but not absolutely convergent.

Emma Aldridge · Answer

undefined To determine the convergence of the series $ \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+4}} $ from $ n = 1 $ to infinity:

1. **Absolute Convergence**: 
   Test if the absolute value of the series converges. If $ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{\sqrt{n+4}} \right| $ converges, the series is absolutely convergent.

2. **Conditional Convergence**: 
   If the series converges but the absolute value of the series diverges, then the series is conditionally convergent.

3. **Divergence**: 
   If neither the original series nor its absolute value converges, the series diverges.

To determine absolute convergence, consider $ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{\sqrt{n+4}} \right| $ and evaluate whether it converges or diverges. If it converges, the original series converges absolutely. If it diverges, further analysis is needed to determine conditional convergence or divergence.