∵an & gt； Bn>0, an & gtA(n+ 1), and the sequence {an} monotonically decreases.
∫lim(n->； ∞)an=0，
According to Leibniz discriminant method of staggered series,
Convergence of staggered series ∑ (n =1..∞) (-1) n * an).
There is a comparison method for positive sequence,
Whether the interleaving sequence can be based on a & gtbn & gt0,
Judge the convergence of ∑ (n = 1) ..∞) (-1) n * bn)? The conclusion is not necessarily.
Example 1. Let an = 1/n and bn =1/(n1+1), which obviously satisfies the condition an >；; Bn>0 and a & gtan+ 1, lim(n->∞)an=0, and bn & gtbn+ 1, lim (n->; ∞)bn=0 holds,
According to Leibniz discriminant method of staggered series, staggered series ∑ (n =1..∞) (-1) n * an) converges, staggered series ∑ (n = 1) (- 1).
Example 2. Take an =1√ n, bn =1(n+1) (n =1,3,5,7, ...) or = 1/(n 2+655).
Obviously, the conditions an & gtbn & gt0 and a & gtan+ 1, lim(n->∞)an=0 are satisfied. According to the Leibniz discriminant method of staggered series, staggered series ∑ (n =1) n * an).
The sum of its first 2n terms.
s2n =- 1/2+ 1/5- 1/4+ 1/ 17...- 1/(2n- 1+ 1)+ 1/((2n)^2+ 1)
=-∑(k= 1，2，..n)( 1/(2k))+∑(k= 1，2，..n)( 1/(4k^2+ 1))
=- 1/2*∑(k= 1，2，..n)( 1/k)+∑(k= 1，2，..n)( 1/(4n^2+ 1))
∫ series-1/2 * ∑ (k =1... ∞) (1/k)-> -∞, and series ∑ (k =1... ∞) (1)
∴lim(n->； ∞)S2n=-∞，
Therefore, the staggered series ∑ (n =1..∞) (-1) n * bn) diverges.
Extended data:
This is a divergent series. You should ask staggered series (-1) n * 1/n, and Leibniz discriminant method is applied to staggered series. Its content is two conditions. First, after removing the sign term, when n tends to infinity, the general term of the series tends to zero, and the signs of all adjacent terms of the series are staggered. Convergence of series satisfying two conditions.
Leibniz discriminant method is available in advanced mathematics textbooks. If you want to know more methods, you can refer to the teaching materials of mathematical analysis, Abel discriminant, Dirichlet discriminant and Rabe discriminant.
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