Determine the Limit of the Monotonic Sequence using the Convergence Definition:$$a_n \equiv \frac{2n+1}{n^2+3}$$

Convergence Definition: The sequence s is said to converge to the number L provided that is $\epsilon \gt0$ then thereis a number N such that:

$n \gt N$ Implies $|s_n -L|\lt \epsilon$

I know the limit is zero but I am not use how to conclude that using the Convergence definition except by this method:

$$lim_n a_n = 0$$ Thus Given $\epsilon \gt 0$ then $\exists N\ni: n\gt N \implies|a_n -0|\lt \epsilon$

Then we write $|\frac{2n+1}{n^2+3} -0| \lt \epsilon$ or $lim_n \frac{2n+1}{n^2+3} = 0$

Then we Divide everything on LHS by $n^2$ and we achieve :

$$lim_n \frac{\frac{2}{n}+\frac{1}{n^2}}{1+\frac{3}{n^2}} = 0$$

Now based on previous proofs we know $\frac {1}{n}$ and $\frac {1}{n^2}$ go to zero when $n\rightarrow \infty$

Thus this becomes $\frac {0}{1}$ leaving us with $0=0$ which is true, concluding that is converges and thus zero is the limit of $a_n$

But I do not think I am allowed to use the algebraic rules for limits which is what I do towards the end of my proof.

So is there a proof version that uses ONLY the Convergence Definition?

  • $\begingroup$ Monotonicity is offtopic here. $\endgroup$ – Did Feb 1 '16 at 7:22

$a_n =\dfrac{2n+1}{n^2+3} \le \dfrac{2n+1}{n^2} \le \dfrac{2}{n}+\dfrac{1}{n^2} \le \dfrac{3}{n^2}$

For, $\epsilon >0$ choose $N=\left[\sqrt{\dfrac{3}{\epsilon}}\right]+1$, and this works, i.e, $|a_n| < \epsilon~\forall~n>N$.

  • $\begingroup$ Why is it $N=\left[\sqrt{\frac{3}{\epsilon}}\right]+1$ and not just $N=\left[\sqrt{\frac{3}{\epsilon}}\right]$ $\endgroup$ – B ry Dec 4 '15 at 23:59
  • $\begingroup$ It doesn't matter, isn't it? $\endgroup$ – Mahbub Alam Dec 5 '15 at 0:00
  • $\begingroup$ well this was one of my answers and i only got $N=\left[\sqrt{\frac{3}{\epsilon}}\right]$ so if I put that as the answer it is correct right? $\endgroup$ – B ry Dec 5 '15 at 0:01
  • $\begingroup$ Yeah the also everything will be alright I think. $\endgroup$ – Mahbub Alam Dec 5 '15 at 0:02
  • $\begingroup$ Even simpler: $3/n^2 \leq 3/n$ for all positive integers $n$. Hence given $\epsilon > 0$ choose $N = \lfloor 3/\epsilon \rfloor$ $\endgroup$ – Simon S Dec 5 '15 at 4:59

We claim that the limit is zero.

If $n \geq 1$, then $$ \frac{2n+1}{n^{2}+3} < \frac{2n+1}{n^{2}} = \frac{2}{n} + \frac{1}{n^{2}}; $$ given any $\varepsilon > 0$, we have $2/n < \varepsilon/2$ if in addition $n > 4/\varepsilon$ and we have $1/n^{2} < \varepsilon/2$ if in addition $n > \sqrt{2/\varepsilon}$. Hence, for every $\varepsilon > 0$, if $n > \max \{ 4/\varepsilon, \sqrt{2/\varepsilon} \}$ then $$ \frac{2n+1}{n^{2}+3} < \varepsilon. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.