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

Move on to Example 1