Proving that the sequence $\lim_{n\to\infty}\frac{2n+3}{n+1}=2$ is Cauchy using the definition of a Cauchy Sequence.

Question

Not really sure how to compute his due to the limit of two $$\lim_{n\to\infty}\frac{2n+3}{n+1}=2$$ Please help. Thanks in advance.

Answer 1

Fix $\varepsilon>0$. Then define $$ f(x)=\frac{2x+3}{x+1}=2+\frac{1}{x+1}. $$

Finally, the sequence $(f(n))_n$ is Cauchy because $$ f(n)-f(m) = \frac{1}{n+1}-\frac{1}{m+1} \le \frac{1}{\min\{n,m\}}< \varepsilon $$ whenever $n,m$ are integers greater than $1/\varepsilon$.

Answer 2

The only way I can see to do this is to note:

As this is cauchy, for any $\epsilon > 0$ we can find an $M$ where $m,n > M$ imply

$|\frac{2n +3}{n+1} - \frac{2m +3}{m+1}| = |2 + \frac1{n+1} - \frac{2m +3}{m+1}| < \epsilon/2$

If we let $n,m > \max(M, 1/\epsilon/2)$ we have $1/(n+1) < \epsilon/2$.

So we have

$\epsilon > |\frac{2m +3}{m+1} -2 - \frac1{n+1}|+|\frac{1}{n+1}|\ge |\frac{2m +3}{m+1} -2 - \frac1{n+1} + \frac{1}{n+1} |$

$= |\frac{2m +3}{m+1} -2 |$

Which proves for $m > \max(M, \frac 1{\epsilon})$, $|\frac{2m +3}{m+1} -2 |<\epsilon$

So $\frac{2m +3}{m+1}\rightarrow 2$