Answer :
Answer:
[tex]a_k=\left|\dfrac{k^2-3}{k^2+4}\right|;\text{ is nondecreasing for $k>2$}[/tex]
[tex]\lim\limits_{k \to \infty} a_k =1[/tex]
The series does not converge
Step-by-step explanation:
Given ...
[tex]S=\displaystyle\sum\limits_{k=2}^{\infty}{x_k}\\\\x_k=(-1)^k\cdot\dfrac{k^2-3}{k^2+4}\\\\a_k=|x_k|[/tex]
Find
whether S converges.
Solution
The (-1)^k factor has a magnitude of 1, so the magnitude of term k can be written as ...
[tex]\boxed{a_k=1-\dfrac{7}{k^2+4}}[/tex]
This is non-decreasing for k>1 (all k-values of interest)
As k gets large, the fraction tends toward zero, so we have ...
[tex]\boxed{\lim\limits_{k\to\infty}{a_k}=1}[/tex]
Terms of the sum alternate sign, approaching a difference of 1. The series does not converge.