Answer:
c
Step-by-step explanation:
Given that:
[tex]\sum \limits ^{\infty}_{k=1} \dfrac{kcos (k\pi)}{k^3+2}[/tex]
since cos (kπ) = [tex]-1^k[/tex]
Then, the series can be expressed as:
[tex]\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^kk)}{k^3+2}[/tex]
In the sum of an alternating series, the best bound on the remainder for the approximation is related to its [tex](n+1)^{th[/tex] term.
∴
[tex]\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(3+1)}(3+1))}{(3+1)^3+2}[/tex]
[tex]\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(4)}(4))}{(4)^3+2}[/tex]
[tex]= \dfrac{4}{64+2}[/tex]
[tex]=\dfrac{2}{33}[/tex]
