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

2 Answers
Feb 9, 2018

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.

Feb 9, 2018

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