How do you determine the convergence or divergence of #sum_(n=1)^(oo) (-1)^(n+1)/n#?

Question

Scarlett Allison · Accepted Answer

The series is convergent and its sum is:

#sum_(n=1)^oo (-1)^(n+1)/n = ln2# Leibniz's theorem states that a sufficient condition for the series with alternating signs: #sum_(n=1)^oo (-1)^n a_n# to be convergent is that: (i) #a_(n+1) <= a_n# (ii) #lim_(n->oo) a_n = 0# Given: #a_n = 1/n# both conditions are satisfied so the series is convergent. We can actually calculate its sum starting from the MacLaurin development of the function #ln(1+x)# In fact: #ln(1+x)|_(x=0) = 0# #d/(dx) (ln(1+x)) = 1/(x+1) =(x+1)^(-1)# #d^2/(dx^2) (ln(1+x)) = -(x+1)^(-2)# #d^3/(dx^3) (ln(1+x)) = 2(x+1)^-3# and we can easily conclude that: #d^n/(dx^n) (ln(1+x)) =( -1)^(n-1)(n-1)! (x+1)^(-n)# and for #x=0# #d^n/(dx^n) (ln(1+x))|_(x=0) = ( -1)^(n-1)(n-1)! (1^(-n)) = ( -1)^(n-1)(n-1)!# so that the MacLaurin series is: #ln(1+x) = sum_(n=1)^oo (-1)^(n-1)(n-1)! x^n/(n!) = sum_(n=1)^oo (-1)^(n-1) x^n/n# As #(-1)^(n-1) = (-1)^(n+1) # if we substitute #x=1# we have: #ln2 = sum_(n=1)^oo (-1)^(n+1)/n#

Emilia Aguilar · Answer

undefined To determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} $, we can use the Alternating Series Test:

1. Verify that the sequence $ \frac{1}{n} $ is positive, decreasing, and approaches zero as $ n $ approaches infinity.
2. Confirm that the terms alternate in sign.

Since both conditions are met, the Alternating Series Test states that if a series $ \sum_{n=1}^{\infty} (-1)^{n+1} a_n $ satisfies the following conditions:

1. $ a_n $ is positive, decreasing, and approaches zero as $ n $ approaches infinity.
2. The terms alternate in sign.

Then the series converges.

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