I need to prove that the following sequence converges:
$\lim_{n\rightarrow \infty} \frac{2n^2+3n+1}{n^2+n+1}=2$
So for the proof/solution I have the following:
Let $\epsilon >0$. Then let $N=\frac{1}{\epsilon}$. Then for all $n\geq N$, $|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
Thus the sequence converges to 2.
Is this the correct way of going about this? Thanks in advance.
$\endgroup$ 12 Answers
$\begingroup$The calculations are right, but some wording changes would be useful. We are given an $\epsilon \gt 0$.
We have, for $n \ge 1$, the inequality $$\left|\frac{2n^2+3n+1}{n^2+n+1}-2\right| \lt \frac{1}{n}.$$ Thus if we put $N=\lceil \frac{1}{\epsilon}\rceil$, then $$\left|\frac{2n^2+3n+1}{n^2+n+1}-2\right|\lt \epsilon$$ for every $n\gt N$.
$\endgroup$ $\begingroup$There is a small error with symbols, it should be like this:
Choose $\epsilon\gt 0$
Let $N=\lceil\frac{1}{\epsilon}\rceil$. Then for all $n\gt N$, $|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
$\endgroup$ 1
