How do you test the series #Sigma sqrt(n+1)-sqrtn# from n is #[0,oo)# for convergence?

1 Answer
Andrea S.
Feb 21, 2017

The series:

#sum_(n=0)^oo sqrt(n+1)-sqrt(n)#

is divergent.

Explanation:

We can see that the series is divergent by analyzing the partial sums:

#s_n = sum_(k=0)^n = (cancel1-0) + (cancel(sqrt2)-cancel1) + (sqrt3-cancel(sqrt2))+...+(cancel(sqrt(n))-sqrt(n-1)) + (sqrt(n+1) -cancel(sqrt(n))) = sqrt(n+1)#

so that:

#lim_(n->oo) s_n = lim_(n->oo) sqrt(n+1) = oo#

We can however use a convergence test in the following way: we have that:

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

Using the identity:

#(a+b)(a-b) = (a^2-b^2)#

this becomes:

#sqrt(n+1)-sqrt(n) = (canceln+1-canceln)/(sqrt(n+1)+sqrt(n)) = 1/(sqrt(n+1)+sqrt(n))#

Now we can use the limit comparison test with the harmonic series:

#lim_(n->oo) (1/(sqrt(n+1)+sqrt(n)))/(1/n) = lim_(n->oo) n/(sqrt(n+1)+sqrt(n)) = oo#

which proves that:

#sum_(n=0)^oo sqrt(n+1)-sqrt(n) = oo#

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