Determine if the series converges or diverges. If the series converges, find its sum? * Both sums are from 1 to infinity*

A) #sum (3)/(n(n+3))#

B) #sum ((1)/(sqrt(n+1)) -(1)/(sqrt(n+3)))#

1 Answer
George C.
May 2, 2018

A) #sum_(n=1)^oo 3/(n(n+3)) = 11/6#

B) #sum_(n=1)^oo (1/sqrt(n+1) - 1/sqrt(n+3)) = sqrt(2)/2+sqrt(3)/3#

Explanation:

A) #sum 3/(n(n+3))#

Note that:

#3/(n(n+3)) = 1/n-1/(n+3)#

So:

#sum_(n=1)^N 3/(n(n+3))#

#= sum_(n=1)^N (1/n-1/(n+3))#

#= sum_(n=1)^N 1/n-sum_(n=4)^(N+3) 1/n#

#= 1/1+1/2+1/3+color(red)(cancel(color(black)(sum_(n=4)^N 1/n)))-color(red)(cancel(color(black)(sum_(n=4)^N 1/n)))-1/(N+1)-1/(N+2)-1/(N+3)#

#= 11/6-1/(N+1)-1/(N+2)-1/(N+3)#

Hence:

#sum_(n=1)^oo 3/(n(n+3)) = lim_(N->oo) sum_(n=1)^N 3/(n(n+3)) = 11/6#

B) #sum (1/sqrt(n+1) - 1/sqrt(n+3))#

#sum_(n=1)^N (1/sqrt(n+1) - 1/sqrt(n+3))#

#= sum_(n=1)^N 1/sqrt(n+1) - sum_(n=1)^N 1/sqrt(n+3)#

#= sum_(n=1)^N 1/sqrt(n+1) - sum_(n=3)^(N+2) 1/sqrt(n+1)#

#= 1/sqrt(2)+1/sqrt(3)+color(red)(cancel(color(black)(sum_(n=3)^N 1/sqrt(n+1)))) - color(red)(cancel(color(black)(sum_(n=3)^N 1/sqrt(n+1))))-1/sqrt(N+2)-1/sqrt(N+3)#

#= sqrt(2)/2+sqrt(3)/3-1/sqrt(N+2)-1/sqrt(N+3)#

Hence:

#sum_(n=1)^oo (1/sqrt(n+1) - 1/sqrt(n+3))#

#= lim_(N->oo) sum_(n=1)^N (1/sqrt(n+1) - 1/sqrt(n+3))#

#= lim_(N->oo) (sqrt(2)/2+sqrt(3)/3-1/sqrt(N+2)-1/sqrt(N+3))#

#= sqrt(2)/2+sqrt(3)/3#

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