How do you test for convergence given #Sigma (-1)^n(1-1/n^2)# from #n=[1,oo)#?

Question

Paisley Johnson · Accepted Answer

The series:

#sum_(n=1)^oo (-1)^n (1-1/n^2)#

is not convergent.

A necessary condition for any series to converge is that the general term of the succession is infinitesimal, that is: #lim_(n->oo) a_n = 0# In our case we have: #lim_(n->oo) (1-1/n^2) = 1# and so: #lim_(n->oo) a_n = lim_(n->oo) (-1)^n(1-1/n^2)# is indeterminate.

Samantha Adkison · Answer

undefined To test for convergence of the series $\sum_{n=1}^{\infty} (-1)^n(1-\frac{1}{n^2})$, you can use the alternating series test. This test states that if the series alternates in sign and the absolute values of the terms decrease monotonically to zero, then the series converges.

In this series, the terms alternate in sign ($(-1)^n$) and the absolute value of each term $(1-\frac{1}{n^2})$ decreases monotonically to zero as $n$ increases. Therefore, according to the alternating series test, the series converges.