# Using the definition of convergence, how do you prove that the sequence lim 2/(sqrt(n+3))=0 converges?

Sep 2, 2017

Given any real number $\epsilon > 0$, we can choose an integer ${N}_{\epsilon}$ such that:

$\frac{{N}_{\epsilon} + 3}{4} > \frac{1}{\epsilon} ^ 2$

Then for $n > {N}_{\epsilon}$ we have:

$\frac{2}{\sqrt{n + 3}} < \frac{2}{\sqrt{{N}_{\epsilon} + 3}}$

$\frac{2}{\sqrt{n + 3}} < \frac{1}{\sqrt{\frac{{N}_{\epsilon} + 3}{4}}}$

$\frac{2}{\sqrt{n + 3}} < \frac{1}{\sqrt{\frac{1}{\epsilon} ^ 2}}$

$\frac{2}{\sqrt{n + 3}} < \epsilon$

which proves the point.