- #1
ActionPotential
- 60
- 4
Homework Statement
Determine the convergence or divergence of the sequence with the given [itex]a_n[/itex] . If it converges, find the limit.
[itex]\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)} [/itex]
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.
[itex]\displaystyle a_n = \frac{n^2}{(2n+1)} - \frac{n^2}{(2n-1)} [/itex]
[itex] \displaystyle a_n = \frac{n}{(2+\frac{1}{n})} - \frac{n}{(2-\frac{1}{n})} [/itex]
[itex] \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})}
= \displaystyle\frac{\infty}{2} - \displaystyle\frac{\infty}{2} : Divergent[/itex]
Alternatively, I noticed if I factored out an n^2 from the top and bottom, and took the limit I would be left with:
[itex] \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[/itex]
However, neither of these approaches seems correct to me.
Thanks for your help and feedback!
Regards,
AP