# Thread: Limit of sequence

1. ## Limit of sequence

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

2. Originally Posted by Len

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 $\displaystyle a_n$.

-Dan

3. 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 $\displaystyle a_n$.

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

4. You want to prove $\displaystyle a_n \to 1$.

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

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

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

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

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

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

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

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

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

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

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

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.