Need help with proving that $(6n^2 -1/ (2n^2 + 5n))$ is a Cauchy Sequence by using definition

Question

Using the definition of the Cauchy Sequence, show that $\forall\varepsilon\gt 0, \exists n_0\in\mathbb N$ s.t. $\forall m\gt n\ge n_0$, $\vert x_m - x_n\vert\le\varepsilon$.

I have tried to estimate it from above and got $\dfrac{12m^2}{4m^2}$, but in that case, the $m$ terms in the numerator and the denominator get canceled.

Answer 1

For $$ x_n = \frac{6n^2-1}{2n^2+5} = \frac{3(2n^2+5)-16}{2n^2+5} = 3 - \frac{16}{2n^2+5}$$ we have, for $m>n\ge n_0$ $$ |x_n-x_m| = \left|\frac{-16}{2n^2+5} - \frac{-16}{2m^2+5}\right| = \frac{32(m^2-n^2)}{(2n^2+5)(2m^2+5)} $$ using $$ m^2-n^2 < m^2 $$ $$ 2m^2+5 > 2m^2 $$ $$ 2n^2+5 > 2n^2 \ge 2n_0^2 $$ we get then $$ |x_n-x_m| < \frac{32m^2}{(2n^2+5)2m^2} = \frac{16}{2n^2+5} < \frac{8}{n_0^2}$$ Therefore, for any $\epsilon>0$, if we take $n_0 >\sqrt{8/\epsilon}$ we have $$ |x_n-x_m| < \frac{8}{(\sqrt{8/\epsilon})^2} = \epsilon$$

Answer 2

well, the definition.

For any $\epsilon > 0$ there is an $N$ so that for any $n,m > N$ then

$|\frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}| < \epsilon$.

And $ \frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}=$

$\frac {6n^2 + 15n -15n-1}{2n^2 + 5n} - {6m^2 + 15m -15m-1}{2m^2 + 5m}=$

$3-\frac {15n+1}{2n^2 + 5n} - 3 + \frac {15m+1}{2m^2 + 5m}=$

$-\frac {15}{2n+5}-\frac 1{2n^2 + 5n}-\frac {15}{2m+5} - \frac 1{2n^2 + 5n}$.

Now we do know that each of those terms converge to $0$.

Or by definition for every $\epsilon_i$ there are $N_i$ so that $n > N_1$ $|-\frac {15}{2n + 5}| < \epsilon_1$ and so on....

If we let $\epsilon_i = \frac \epsilon 4$ and $N = \max (N_i)$ then if $n,m > M$ then

$|-\frac {15}{2n+5}-\frac 1{2n^2 + 5n}-\frac {15}{2m+5} - \frac 1{2n^2 + 5n}|=$

$|\frac {15}{2n+5}|+|\frac 1{2n^2 + 5n}|+|\frac {15}{2m+5}|+| + \frac 1{2n^2 + 5n}| <$

$4\cdot \frac \epsilon 4 < \epsilon$.

If need be we can caculate and find these $N_i$ individually.

we need $\frac {15}{2n + 5} < \frac \epsilon 4$ so

$\frac {\frac {15}{\frac \epsilon 4}-5}2 < n$ so we can have $N_1=N_3 = \frac {30}\epsilon$

we need $\frac 1{2n^2 + 5n} < \frac {\epsilon}4$ so $\frac 4{\epsilon} < 2n^2 + 5n$ so we can have $\frac 2{\epsilon} < n^2 < n^2 - \frac 52n$ if $N_2 = N_4 =\sqrt{\frac 2{\epsilon}}$

So if $N > \max (\frac {30}\epsilon, \sqrt{\frac 2{\epsilon}})$ we will have $n,m > N$ would imply:

$|\frac{6n^2 -1}{2n^2 + 5n} - \frac{6m^2 -1}{2m^2 + 5m}| =$

$|\frac {15}{2n+5}+\frac 1{2n^2 + 5n}+\frac {15}{2m+5} + \frac 1{2n^2 + 5n}|=$

$|\frac {15}{2n+5}|+|\frac 1{2n^2 + 5n}|+|\frac {15}{2m+5}|+| \frac 1{2n^2 + 5n}|<$

$|\frac {15}{2n}| + |\frac 1{2n^2}| + |\frac {15}{2m}| + |\frac 1{2n^2}| <$

$|\frac {15}{2\frac {30}\epsilon}| +|\frac 1{2\frac 2{\epsilon}}|+|\frac {15}{2\frac {30}\epsilon}| +|\frac 1{2\frac 2{\epsilon}}|=$

$|\frac \epsilon 4| +|\frac \epsilon 4| +|\frac \epsilon 4| +|\frac \epsilon 4| =\epsilon$