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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?

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  • $\begingroup$ Monotonicity is offtopic here. $\endgroup$ – Did Feb 1 '16 at 7:22
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$a_n =\dfrac{2n+1}{n^2+3} \le \dfrac{2n+1}{n^2} \le \dfrac{2}{n}+\dfrac{1}{n^2} \le \dfrac{3}{n^2}$

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

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  • $\begingroup$ 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
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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 $$ \frac{2n+1}{n^{2}+3} < \varepsilon. $$

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