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{√(n+2)}-\frac3{√(n+3)}`

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

`S_k=\displaystyle\sum_(n=1)^k\frac3{√(n+2)}-\frac3{√(n+3)}`

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

`\implies S_k=\frac3{\sqrt3}-\frac3{√(k+3)}`

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

`\displaystyle\sum_(n=1)^\infty\frac3{√(n+2)}-\frac3{√(n+3)}=\frac3{\sqrt3}=\boxed{\sqrt3}}`