How do you use the integral test to determine if #Sigma1/sqrt(n+1)# from #[1,oo)# is convergent or divergent?

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Nov 14, 2017

Answer:

The series is divergent

Explanation:

Apply the integral test only if the function #f(x)# is continuous, positive and decreasing.

The function #f(x)# is continuous as #AA x in [1,oo)# as #sqrt(1+x)# is continuous.

The function is positive as #sqrt(1+x)# is positive in the interval #[1,oo]#

#f'(x)=(1/sqrt(1+x))'=-1/2(1+x)^(-3/2)#

Therefore,

#f'(x)<0, AA x in [1,oo)#

So, we can apply the integral test.

Calculate the improper integral

#int_1^oodx/sqrt(1+x)#

First compute the indefinite integral

#intdx/sqrt(1+x)=int(1+x)^(-1/2)dx#

#=(1+x)^(-1/2+1)/(-1/2+1)#

#=2(1+x)^(1/2)#

Therefore,

#lim_(t->oo)int_1^tdx/sqrt(1+x)=lim_(t->oo)[2(1+x)^(1/2)]_1^t#

#=lim_(t->oo)(2sqrt(1+t)-2sqrt2)#

#=+oo#

As the improper integral is divergent , so the #sum_(n=1)^oo1/sqrt(1+n)# is divergent

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