# Limit of sequence

• Mar 12th 2008, 03:28 PM
Len
Limit of sequence
http://img145.imageshack.us/img145/387/inflimit2ej5.jpg

Using n>=N ==> | A_n - L | < e

Most the proofs I read say to start with | a_n - 1 | < e and work backwards to find N but I have done so and I cannot eliminate then n term. I am stuck and any tips / help would be hugely appreciated.

I end up getting

1+2n - (-1)^n
-------------- < e
n^2 +2n+1
• Mar 12th 2008, 03:30 PM
topsquark
Quote:

Originally Posted by Len
http://img208.imageshack.us/img208/9058/inflimitvf1.jpg

Most the proofs I read say to start with | a_n - 1 | < e and work backwards to find N but I have done so and I cannot eliminate then n term. I am stuck and any tips / help would be hugely appreciated.

I end up getting

1+2n - (-1)^n
-------------- < e
n^2 +2n+1

Your sequence definition makes no sense. The left hand side is independent of n, but you are using that to define your $a_n$.

-Dan
• Mar 12th 2008, 03:32 PM
Len
Quote:

Originally Posted by topsquark
Your sequence definition makes no sense. The left hand side is independent of n, but you are using that to define your $a_n$.

-Dan

Sorry I wrote that wrong. I am trying to show that A_n converges to 1.

http://img145.imageshack.us/img145/387/inflimit2ej5.jpg
• Mar 12th 2008, 03:45 PM
ThePerfectHacker
You want to prove $a_n \to 1$.

Thus, we have to show $\left| \frac{n^2+(-1)^n}{(n+1)^2} - 1 \right| < \epsilon$.

Thus, $\left| \frac{n^2 + (-1)^n - (n+1)^2}{(n+1)^2}\right| < \epsilon$.

Thus, $\left| \frac{n^2 + (-1)^n - n^2 - 2n - 1}{(n+1)^2} \right| < \epsilon \implies \left| \frac{(-1)^n - 2n}{(n+1)^2} \right| < \epsilon$.

But note, $\left| \frac{(-1)^n - 2n }{(n+1)^2} \right| \leq \frac{2n+1}{n^2} \leq \frac{2n + n}{n^2} \leq \frac{3}{n}$.

So if we can make $3/n < \epsilon$ then $|a_n - 1| < \epsilon$. And this of course can be made by choosing $N > 3/\epsilon$.
• Mar 12th 2008, 04:16 PM
Len
Quote:

Originally Posted by ThePerfectHacker
You want to prove $a_n \to 1$.

Thus, we have to show $\left| \frac{n^2+(-1)^n}{(n+1)^2} - 1 \right| < \epsilon$.

Thus, $\left| \frac{n^2 + (-1)^n - (n+1)^2}{(n+1)^2}\right| < \epsilon$.

Thus, $\left| \frac{n^2 + (-1)^n - n^2 - 2n - 1}{(n+1)^2} \right| < \epsilon \implies \left| \frac{(-1)^n - 2n}{(n+1)^2} \right| < \epsilon$.

But note, $\left| \frac{(-1)^n - 2n }{(n+1)^2} \right| \leq \frac{2n+1}{n^2} \leq \frac{2n + n}{n^2} \leq \frac{3}{n}$.

So if we can make $3/n < \epsilon$ then $|a_n - 1| < \epsilon$. And this of course can be made by choosing $N > 3/\epsilon$.

http://img146.imageshack.us/img146/9...flimit3xk9.jpg

Thanks for help but my last question is do i need the = between

|a_n - 1| <= 3/n . I do not think they are ever equal.