$\displaystyle
\sum_{n=1}^\infty \frac{2}{sqrt (n^2+3)}
$
I tryed to use the integral test
came up with 2/2(n^2+3)^3/2) from 1 to infinity
0-1/8 = -1/8 ?
$\displaystyle
\sum_{n=1}^\infty \frac{2}{sqrt (n^2+3)}
$
I tryed to use the integral test
came up with 2/2(n^2+3)^3/2) from 1 to infinity
0-1/8 = -1/8 ?
Use the comparison test. Note that $\displaystyle \frac{2}{\sqrt{n^2 + 3}} \geq \frac{2}{\sqrt{n^2 + 3n^2}} = \frac{1}{n}$ for $\displaystyle n \geq 1$.