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#