How do you test the alternating series #Sigma (n(-1)^(n+1))/lnn# from n is #[2,oo)# for convergence?

Question

Charles Arocha · Accepted Answer

The series:

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

is not convergent.

A necessary condition for any series to converge is that: #lim_(n->oo) a_n = 0# which also implies: #lim_(n->oo) abs(a_n) = 0# In our case: #lim_(n->oo) abs(a_n) = lim_(n->oo) n/lnn = oo# so the series is not convergent. We can also see that taking only the terms of even order: #lim_(n->oo) a_(2n) = lim_(n->oo) (2n(-1)^(2n+1))/ln(2n) = lim_(n->oo) -(2n)/ln(2n) = -oo# while for terms of odd order: #lim_(n->oo) a_(2n+1) = lim_(n->oo) ((2n+1)(-1)^(2n+2))/ln(2n+1) = lim_(n->oo) (2n+1)/ln(2n+1) = +oo# so the series is irregular.

Ezekiel Alexander · Answer

undefined To test the alternating series $ \sum_{n=2}^{\infty} \frac{n(-1)^{n+1}}{\ln n} $ for convergence, you can use the Alternating Series Test. Here's how:

1. Verify if the series satisfies the conditions of the Alternating Series Test:
   a. The terms of the series alternate in sign.
   b. The absolute values of the terms decrease as $ n $ increases.
   c. The limit of the absolute values of the terms approaches zero as $ n $ approaches infinity.

2. Check if the terms alternate in sign: The series alternates between positive and negative terms as $ n $ changes.

3. Examine if the absolute values of the terms decrease: Consider the absolute value of each term $ \frac{n(-1)^{n+1}}{\ln n} $. Confirm whether these absolute values decrease as $ n $ increases.

4. Evaluate the limit of the absolute values of the terms: Calculate $ \lim_{n \to \infty} \frac{n}{\ln n} $. If this limit equals zero, the series satisfies the conditions of the Alternating Series Test.

5. If all conditions of the Alternating Series Test are met, then the series converges. Otherwise, it diverges.