Question

Determine whether the series $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n^2+3n+6}}$ converges or diverges?

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

Answer 1

Answer: (A)

Always diverges

Given: ${\sum }_{n=1}^{\infty }\frac{1}{\sqrt{(n^2+3n+6)}}$

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

${\sum }_{n=1}^{\infty }\frac{1}{\sqrt{(n^2+3n+6)}}=\infty$ Hence series diverges always.