How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n+1)n^2)/(n+1)^2# from #[1,oo)#?

Question

Emilia Aguilar · Accepted Answer

The series diverges.

The series #sum_(n=1)^oo(-1)^(n+1)n^2/(n+1)^2# is made up of two parts. The #(-1)^(n+1)# part alternates between #1# and #-1#, only changing the sign of each term. The meat of the sequence is #n^2/(n+1)^2#. Note that #lim_(nrarroo)n^2/(n+1)^2=1#. So, as the series extends infinitely, alternating terms approaching a value of #1# are continually being added and subtracted, so the series never converges.

Charles Arocha · Answer

undefined To determine the convergence properties of the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} $ from $ n = 1 $ to infinity, we'll apply the alternating series test and the ratio test.

1. Alternating Series Test:
   - The series alternates in sign and the absolute values of terms decrease monotonically to zero as $ n $ increases.
   - Thus, the series converges conditionally.

2. Ratio Test:
   - Taking the limit as $ n $ approaches infinity of the absolute value of the ratio of consecutive terms:
     $$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+2} (n+1)^2}{(n+2)^2}}{\frac{(-1)^{n+1} n^2}{(n+1)^2}} \right| $$
     $$ = \lim_{n \to \infty} \left| \frac{(n+1)^2}{(n+2)^2} \right| = 1 $$
   - Since the limit is 1, the test is inconclusive.

Therefore, the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} $ converges conditionally.