How do you determine the convergence or divergence of #Sigma ((-1)^(n))/(ln(n+1))# from #[1,oo)#?

Question

Isaiah Allen · Accepted Answer

The series:

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

is convergent.

The series:

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

is convergent.

The series:

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

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

#sum_(n=1)^oo (-1)^n a_n#

is convergent if:

(i) #lim_(n->oo) a_n = 0#

(ii) #a_(n+1) <= a_n#

in our case:

#lim_(n->oo) 1/ln(n+1) = 0#

and since #lnx# is a monotone growing function:

#ln n < ln(n+1) <=> 1/lnn > 1/ln(n+1)#

also the second condition is satisfied.

Paisley Johnson · Answer

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

1. Check if the sequence $ \frac{1}{\ln(n+1)} $ is positive, decreasing, and approaches zero as $ n $ approaches infinity.
2. Verify if it satisfies the conditions of the Alternating Series Test.

For the Alternating Series Test:
- The terms alternate in sign.
- The absolute value of the terms decreases as $ n $ increases.
- The terms approach zero as $ n $ approaches infinity.

1. The sequence $ \frac{1}{\ln(n+1)} $ is positive for all $ n \geq 1 $.
2. $ \frac{1}{\ln(n+1)} $ decreases as $ n $ increases because $ \ln(n+1) $ increases as $ n $ increases.
3. As $ n $ approaches infinity, $ \frac{1}{\ln(n+1)} $ approaches zero.

Since the sequence $ \frac{1}{\ln(n+1)} $ satisfies all the conditions of the Alternating Series Test, the series $ \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)} $ converges.