This is an interactive website to help you learn how to write an $$\varepsilon - K$$ proof. How would you address this problem?

### Prove that $$\left(\tfrac{2}{n+3}\right)\rightarrow 0$$.

Where do you start? As always, you study an appropriate definition! Let's examine the definition of sequence convergence.

Let $$(x_n)$$ be a sequence of real numbers. A real number $$x$$ is said to a limit of $$(x_n)$$, or $$(x_n)$$ is said to converge to $$x$$, if for every $$\varepsilon>0$$ there is a number $$K$$ such that for all $$n\ge K$$ the terms in $$(x_n)$$ satisfy $$|x_n-x| < \varepsilon$$.

What does this say? It's really a very complex sentence. For any positive number $$\varepsilon > 0$$ that is given to us (we have no control over it), we need to find a number $$K$$ (which will be a function of $$\varepsilon$$, not $$n$$) with the property that if $$n$$ is any number greater than or equal to $$K$$, we have $$|x_n-x| < \varepsilon$$.

The bottom line: We need $$|x_n-x| < \varepsilon$$! Let's do a little scratch work to see what it will take for that to be true.

To have $$|x_n-x| < \varepsilon$$, in this problem we need $$\left|\left(\tfrac{2}{n+3}\right)-0\right| < \varepsilon$$. Subtracting $$0$$ is easy to deal with, and $$\left(\tfrac{2}{n+3}\right)$$ is always positive, so we've reduce the problem to making sure that $$\left(\tfrac{2}{n+3}\right) < \varepsilon$$. Solving for $$n$$ (since we need a condition on $$n$$) yields

$$\left(\tfrac{2}{n+3}\right) < \varepsilon \Rightarrow$$
$$\left(\tfrac{2}{\varepsilon}\right) < n+3 \Rightarrow$$
$$\left(\tfrac{2}{\varepsilon}\right)-3 < n.$$

We've discovered that if we let $$K > \left(\tfrac{2}{\varepsilon}\right)-3$$, then $$n\ge K$$ will imply $$\left|\left(\tfrac{2}{n+3}\right)-0\right| < \varepsilon$$, once we write a proper proof!

How does the information we have fit into a proof? Try writing the proof yourself using the proof template, then click on the box below to see a finished proof.

Show the Proof

Proof: Given $$\varepsilon >0$$, let $$K > \left(\tfrac{2}{\varepsilon}\right)-3$$. Then if $$n\ge K$$ we have $$\left|\left(\tfrac{2}{n+3}\right)-0\right| =$$ $$\left(\tfrac{2}{n+3}\right)\le$$ $$\left(\tfrac{2}{K+3}\right)<$$ $$\left(\tfrac{2}{\left(\left(\tfrac{2}{\varepsilon}\right)-3\right)+3}\right)=$$ $$\varepsilon$$. Thus $$\left(\tfrac{2}{n+3}\right)\rightarrow 0$$ by the Definition of Sequence Converence.

The expressions in red arose naturally in the computations, and were used in the proof.