# Limit Comparison Series

• July 24th 2013, 08:08 PM
minneola24
Limit Comparison Series
So if I compare it to 1/n that new comparison diverges.

But 1/n is bigger than the original function so I cant conclude that the original diverges too.

If I divide the original by the 1/n and take the limit as n-> infinity I keep going in loops using lhopitals.

How do I solve this?

Thanks.
• July 24th 2013, 08:24 PM
Prove It
Re: Limit Comparison Series
Quote:

Originally Posted by minneola24
So if I compare it to 1/n that new comparison diverges.

But 1/n is bigger than the original function so I cant conclude that the original diverges too.

If I divide the original by the 1/n and take the limit as n-> infinity I keep going in loops using lhopitals.

How do I solve this?

Thanks.

\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^2 + 4}} &> \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^2 + 4n + 4}} \\ &= \sum_{ n = 1}^{\infty} \frac{1}{\sqrt{(n + 2)^2}} \\ &= \sum_{n = 1}^{\infty} \frac{1}{n + 2} \textrm{ since } n +2 > 0 \textrm{ for all } n \textrm{ in our sum } \\ &= \sum_{n = 3}^{\infty} \frac{1}{n} \end{align*}

which is a harmonic series, and is divergent. Therefore the original series is divergent.
• July 24th 2013, 08:43 PM
Prove It
Re: Limit Comparison Series
Another possibility is the integral test. We can compare the sum to the function \displaystyle \begin{align*} \frac{1}{\sqrt{x^2 + 4}} \end{align*}. Since the function is always decreasing when \displaystyle \begin{align*} x > 0 \end{align*}, a left-hand endpoint estimate of the integral will be an over-estimate. Notice that when we choose to have each of our subintervals be 1 unit in length, then the integral \displaystyle \begin{align*} \int_1^{\infty}{\frac{1}{\sqrt{x^2 + 4}}\,dx} \end{align*} can be approximated by \displaystyle \begin{align*} 1 \cdot \frac{1}{\sqrt{1^2 + 4}} + 1 \cdot \frac{1}{\sqrt{2^2 + 4}} + 1 \cdot \frac{1}{\sqrt{3^2 + 4}} + \dots = \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^2 + 4}} \end{align*}. But since this is an overestimate, we can say

\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^2 + 4}} &> \int_1^{\infty}{\frac{1}{\sqrt{x^2 + 4}}\,dx} \\ &= \int_{\sinh^{-1}{\left( \frac{1}{2} \right) } }^{\infty}{ \frac{2\cosh{(t)}}{\sqrt{\left[ 2\sinh{(t)} \right] ^2 + 4}}\,dt } \textrm{ when we substitute } x = 2\sinh{(t)} \\ &= \int_{\sinh^{-1}{\left( \frac{1}{2} \right) } }^{\infty}{ \frac{2\cosh{(t)}}{\sqrt{4 \left[ \sinh^2{(t)} + 1 \right] }} \,dt } \\ &= \int_{\sinh^{-1}{\left( \frac{1}{2} \right) } }^{\infty}{ \frac{2\cosh{(t)}}{2\cosh{(t)}}\,dt} \\ &= \int_{\sinh^{-1}{\left( \frac{1}{2} \right) } }^{\infty} {1\,dt} \\ &= \lim_{\epsilon \to \infty} \left[ t \right] _{\sinh^{-1}{\left( \frac{1}{2} \right) }}^{\epsilon} \\ &= \lim_{\epsilon \to \infty} \left( \epsilon - \sinh^{-1}{\left( \frac{1}{2} \right) } \right) \\ &\to \infty \end{align*}

Since the sum is greater than this divergent integral, the sum must also diverge.
• July 25th 2013, 04:31 AM
Plato
Re: Limit Comparison Series
Quote:

Originally Posted by minneola24
So if I compare it to 1/n that new comparison diverges.
But 1/n is bigger than the original function so I cant conclude that the original diverges too.
If I divide the original by the 1/n and take the limit as n-> infinity I keep going in loops using lhopitals.

When using the limit comparison test it is best to do direct computations.

Let $a_n=n^{-1}~\&~b_n=\left(\sqrt{n^2+1}\right)^{-1}.$

Now ${\lim _{x \to \infty }}\left( {\frac{{{a_n}}}{{{b_n}}}} \right) = {\lim _{x \to \infty }}\sqrt {1 + \frac{1}{{{n^2}}}} = 1$

Thus we do not need to 'worry' with direct comparison.