Question

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

Answer 1

The series:

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

is divergent.

Explanation:

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.