How do you use the limit comparison test to determine if #Sigma 1/sqrt(n^2+1)# from #[0,oo)# is convergent or divergent?

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Answer 1 · 2017-02-21T21:15:14

The series:

#sum_(n=0)^oo 1/sqrt(n^2+1)#

is divergent.

The series:

#sum_(n=0)^oo a_n = sum_(n=0)^oo 1/sqrt(n^2+1)#

has positive terms #a_n>0#.
The limit comparison test tells us that if we find another series with positive terms:

#sum_(n=0)^oo b_n#

such that:

#lim_(n->oo) a_n/b_n = L# with #L in (0,+oo)#

then the two series are either both convergent or both divergent.

Now, clearly: #sqrt(n^2+1) ~= n#, so we can choose as test series the harmonic series:

#sum_(n=1)^oo 1/n -> oo#

and in fact:

#lim_(n->oo) (1/sqrt(n^2+1))/(1/n) = lim_(n->oo) n/sqrt(n^2+1) = 1#

which proves that the series:

#sum_(n=0)^oo 1/sqrt(n^2+1)#

is divergent.