Question

Determine whether the series $\sum _{n=1}^{\infty }\frac{6n}{(n^2+2{)}^4}$ converges or diverges?

• A. Always diverges
• B. Always converges
• C. Conditionally converges
• D. Conditionally diverges

Answer 1

The correct option is B Always converges

Given : ${\sum }_{n=1}^{\infty }\frac{6n}{{(n^2+2)}^4}$

$\stackrel{lim}{c\to \infty }{\sum }_{n=1}^{\infty }\frac{6n}{{(n^2+2)}^4}=\stackrel{lim}{c\to \infty }{\int }_0^c\frac{6n}{{(n^2+2)}^4}dx=\stackrel{lim}{c\to \infty }{\int }_0^c\frac{2x}{{(x^2+2)}^4}dx{\sum }_{n=1}^{\infty }\frac{6n}{{(n^2+2)}^4}=\stackrel{lim}{c\to \infty }{\left|\frac{{(x^2+2)}^{-3}}{-3}\right|}^c=\stackrel{lim}{c\to \infty }-[{(c^2+2)}^{-3}-{\left(3\right)}^{-3}]{\sum }_{n=1}^{\infty }\frac{6n}{{(n^2+2)}^4}=\stackrel{lim}{c\to \infty }+3^{-3}=\frac{1}{27}$

Hence series always converges.