Determine Convergence/Divergence of Sequence (n^2/(2n+1) - (n^2/2n-1)

[ActionPotential]

Homework Statement

Determine the convergence or divergence of the sequence with the given $a_n$ . If it converges, find the limit.

$\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)}$

The Attempt at a Solution

[/B]
I am not confident enough with sequence and series to know that I have done this correctly so I am hoping someone can let me know if this is correct and if not, what it is that I am doing incorrectly. I also thought about trying to determine whether or not the sequence is monotonic and whether it has an upper and/or lower bound but wasn't sure if this was necessary.

$\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)}$

$\displaystyle a_n = \frac{n}{(2+\frac{1}{n})} - \frac{n}{(2-\frac{1}{n})}$

$\displaystyle\lim_{n\rightarrow +\infty} a_n = \displaystyle\lim_{n\rightarrow +\infty} \frac{n}{(2+\frac{1}{n})} - \displaystyle\lim_{n\rightarrow +\infty} \frac{n}{(2-\frac{1}{n})} <br /> = \displaystyle\frac{\infty}{2} - \displaystyle\frac{\infty}{2} : Divergent$

Alternatively, I noticed if I factored out an n^2 from the top and bottom, and took the limit I would be left with:

$\displaystyle\lim_{n\rightarrow +\infty} a_n = \displaystyle\lim_{n\rightarrow +\infty} \displaystyle \frac{1}{(\frac{2}{n}+\frac{1}{n^2})} - \displaystyle \frac{1}{(\frac{2}{n}-\frac{1}{n^2})} = \frac{1}{0} - \frac{1}{0} : Divergent$

However, neither of these approaches seems correct to me.

Thanks for your help and feedback!

Regards,
AP

 

[Ray Vickson]

Homework Statement

Determine the convergence or divergence of the sequence with the given $a_n$ . If it converges, find the limit.

$\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)}$

The Attempt at a Solution

[/B]
I am not confident enough with sequence and series to know that I have done this correctly so I am hoping someone can let me know if this is correct and if not, what it is that I am doing incorrectly. I also thought about trying to determine whether or not the sequence is monotonic and whether it has an upper and/or lower bound but wasn't sure if this was necessary.

$\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)}$

$\displaystyle a_n = \frac{n}{(2+\frac{1}{n})} - \frac{n}{(2-\frac{1}{n})}$

$\displaystyle\lim_{n\rightarrow +\infty} a_n = \displaystyle\lim_{n\rightarrow +\infty} \frac{n}{(2+\frac{1}{n})} - \displaystyle\lim_{n\rightarrow +\infty} \frac{n}{(2-\frac{1}{n})}<br /> = \displaystyle\frac{\infty}{2} - \displaystyle\frac{\infty}{2} : Divergent$

Alternatively, I noticed if I factored out an n^2 from the top and bottom, and took the limit I would be left with:

$\displaystyle\lim_{n\rightarrow +\infty} a_n = \displaystyle\lim_{n\rightarrow +\infty} \displaystyle \frac{1}{(\frac{2}{n}+\frac{1}{n^2})} - \displaystyle \frac{1}{(\frac{2}{n}-\frac{1}{n^2})} = \frac{1}{0} - \frac{1}{0} : Divergent$

However, neither of these approaches seems correct to me.

Thanks for your help and feedback!

Regards,
AP

Both of these answers/appproaches are wrong. Before even starting you should re-write $a_n$ in a simpler form. Start by re-writing
$$\frac{1}{2n+1}-\frac{1}{2n-1}$$

 

[jfizzix]

I would factor out the n^2, and then simplify the difference of fractions so that they have the same denominator. If that fraction falls to zero more slowly than 1/n^2, then the sequence is divergent. If it isn't divergent, you could use the ratio test to check for convergence.

 

[ActionPotential]

Thanks for your help, Ray. Are you suggesting that I compare $a_n$ with a simpler sequence $b_n$ such as the one you responded with? Sorry for the confusion, I am still learning these approaches/techniques.

 

[ActionPotential]

I had originally typed out an entire post but it got accidentally deleted so I am just skipping straight to the end. I let f(n)=a_n and substituted for f(x). Combined rational expressions and reduced it to a function of like denominator. Not entirely sure if this is correct either.

$\displaystyle f(x) = \frac{x^2(2x-1) - x^2(2x+1)}{(2x+1)(2x-1)} = \frac{2x^3-x^2-2x^3-x^2}{4x^2-1} = \frac{-2x^2}{4x^2-1}$

$\displaystyle\lim_{x\rightarrow +\infty} f(x) = \displaystyle\lim_{x\rightarrow +\infty} \displaystyle \frac{-2x^2}{(4x^2-1)} = -\frac{1}{2}$ therefore $\displaystyle\lim_{x\rightarrow +\infty} a_n = -\frac{1}{2} :Converges$

We can take l'hospital's Rule twice and arrive at -1/2 as well.

 

[Ray Vickson]

Thanks for your help, Ray. Are you suggesting that I compare $a_n$ with a simpler sequence $b_n$ such as the one you responded with? Sorry for the confusion, I am still learning these approaches/techniques.

I said exactly what I think you should do---re-write $a_n$. Don't compare, don't do anything else; just re-write.