Question

Calculate `sum_(n=0)^oo sqrt(n+3)+sqrtn-2sqrt(n+2)` ?

Answer 1

Telescoping Series 1

Explanation:

`Sigma(sqrt(n+2) - 2sqrt(n+1)+sqrt(n))`
`Sigma(sqrt(n+2) - sqrt(n+1)-sqrt(n+1) + sqrt(n))`
`Sigma((sqrt(n+2) - sqrt(n+1))((sqrt(n+2) + sqrt(n+1))/(sqrt(n+2) + sqrt(n+1)))+(-sqrt(n+1) + sqrt(n))((sqrt(n+1) + sqrt(n))/(sqrt(n+1) + sqrt(n))))`
`Sigma(1/(sqrt(n+2) + sqrt(n+1))+(-1)/(sqrt(n+1) + sqrt(n))))`
This is a collapsing (telescoping) series.
Its first term is
`-1/(sqrt(2) + 1)=1-sqrt2`.

Answer 2

See below.

Explanation:

This is equivalent to

`sum_(n=3)^oo sqrtn+sum_(n=1)^oo sqrtn - 2 sum_(n=2)^oo sqrtn = 1-sqrt2`