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

Dec 22, 2016

By comparing to ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2$, we find that ${\sum}_{n = 1}^{\infty} \frac{1}{n \sqrt{{n}^{2} + 1}}$ is convergent through the limit comparison test.

Explanation:

The limit comparison test states that if $L = {\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n}$ where $L$ is a positive, finite value, then $\sum {a}_{n}$ and $\sum {b}_{n}$ either both converge or both diverge.

When working with the limit comparison test, it's often helpful to let the series we are working with be $\sum {a}_{n}$. Thus, let ${a}_{n} = \frac{1}{n \sqrt{{n}^{2} + 1}}$.

The trickier thing to do is choose ${b}_{n}$. The best thing to do is make ${b}_{n}$ mimic whatever ${a}_{n}$ does at infinity.

As $\frac{1}{n \sqrt{{n}^{2} + 1}}$ gets larger and larger, the constant becomes more and more irrelevant and we see that ${a}_{n} \approx \frac{1}{n \sqrt{{n}^{2}}} = \frac{1}{n \left(n\right)} = \frac{1}{n} ^ 2$.

So, let ${b}_{n} = \frac{1}{n} ^ 2$.

Now going to the limit comparison test, let $L = {\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n}$.

Then $L = {\lim}_{n \rightarrow \infty} \frac{\frac{1}{n \sqrt{{n}^{2} + 1}}}{\frac{1}{n} ^ 2} = {\lim}_{n \rightarrow \infty} {n}^{2} / \left(n \sqrt{{n}^{2} + 1}\right)$

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

=lim_(nrarroo)n/(nsqrt(1+1/n^2)

=lim_(nrarroo)1/sqrt(1+1/n^2

$= \frac{1}{\sqrt{1 + 0}}$

$= 1$

So $L = 1$. Since this is positive and finite, we know that $\sum {a}_{n}$ and $\sum {b}_{n}$ have the same convergence or divergence.

We know that $\sum {b}_{n} = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2$ converges through the p-series test, so this means that $\sum {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{1}{n \sqrt{{n}^{2} + 1}}$ also converges through the limit comparison test.