Question

Find the sum of the series Summation from n equals 1 to infinity (StartFraction 3 Over StartRoot n plus 2 EndRoot EndFraction minus StartFraction 3 Over StartRoot n plus 3 EndRoot EndFraction ).

Answer 1

Looks like the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}$

The series telescopes; consider the $k$th partial sum of the series,

$S_k=\displaystyle\sum_{n=1}^k\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}$

$S_k=\displaystyle\left(\frac3{\sqrt3}-\frac32\right)+\left(\frac32-\frac3{\sqrt5}\right)+\cdots+\left(\frac3{\sqrt{k+1}}-\frac3{\sqrt{k+2}}\right)+\left(\frac3{\sqrt{k+2}}-\frac3{\sqrt{k+3}}\right)$

$\implies S_k=\dfrac3{\sqrt3}-\dfrac3{\sqrt{k+3}}$

As $k\to\infty$, the second term converges to 0, leaving us with

$\displaystyle\sum_{n=1}^\infty\frac3{\sqrt{n+2}}-\frac3{\sqrt{n+3}}=\frac3{\sqrt3}=\boxed{\sqrt3}}$