Looks like the series is supposed to be
[tex]\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}[/tex]
The series telescopes; consider the [tex]k[/tex]th partial sum of the series,
[tex]S_k=\displaystyle\sum_{n=1}^k\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}[/tex]
[tex]S_k=\displaystyle\left(\frac3{\sqrt3}-\frac32\right)+\left(\frac32-\frac3{\sqrt5}\right)+\cdots+\left(\frac3{\sqrt{k+1}}-\frac3{\sqrt{k+2}}\right)+\left(\frac3{\sqrt{k+2}}-\frac3{\sqrt{k+3}}\right)[/tex]
[tex]\implies S_k=\dfrac3{\sqrt3}-\dfrac3{\sqrt{k+3}}[/tex]
As [tex]k\to\infty[/tex], the second term converges to 0, leaving us with
[tex]\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}=\frac3{\sqrt3}=\boxed{\sqrt3}}[/tex]
