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

Feb 21, 2017

The series:

${\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{{n}^{2} + 1}}$

is divergent.

#### Explanation:

The series:

${\sum}_{n = 0}^{\infty} {a}_{n} = {\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{{n}^{2} + 1}}$

has positive terms ${a}_{n} > 0$.
The limit comparison test tells us that if we find another series with positive terms:

${\sum}_{n = 0}^{\infty} {b}_{n}$

such that:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = L$ with $L \in \left(0 , + \infty\right)$

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

Now, clearly: $\sqrt{{n}^{2} + 1} \cong n$, so we can choose as test series the harmonic series:

${\sum}_{n = 1}^{\infty} \frac{1}{n} \to \infty$

and in fact:

${\lim}_{n \to \infty} \frac{\frac{1}{\sqrt{{n}^{2} + 1}}}{\frac{1}{n}} = {\lim}_{n \to \infty} \frac{n}{\sqrt{{n}^{2} + 1}} = 1$

which proves that the series:

${\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{{n}^{2} + 1}}$

is divergent.