Question

Prove, using the definition of the limit of a sequence, that \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}}=0\)

Short answer

The limit \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} = 0\) can be proved using the definition of the limit of a sequence by finding a suitable \(N\) such that for all \(n > N\), the term \(\frac{1}{n^3}\) is less than any given positive number \(\epsilon\).

Step-by-step solution

Write down the definition of a limit

To start, recall the mathematical definition of a limit for a sequence. It states that for every \(\epsilon > 0\), there exists some \(N > 0\) such that for all \(n > N\), we have \(|\frac{1}{n^3} - L| < \epsilon\), where \(L\) is the limit of the sequence and \(|\frac{1}{n^3} - L|\) is the absolute value of the difference between the nth term of the sequence and the limit.

Substitute the limit into the definition

Given that the limit \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} = 0\), replace \(L\) with 0 in the definition. This means the inequality becomes \(|\frac{1}{n^3} - 0| < \epsilon\), simplifying to \(|\frac{1}{n^3}| < \epsilon\), or \(\frac{1}{n^3} < \epsilon\) since the sequence is positive.

Solve for N

The inequality can be rewrittten as \(n^3 > \frac{1}{\epsilon}\), taking advantage of the fact that for positive \(x\) and \(y\), if \(x\frac{1}{y}\).

Choose an appropriate N

Finally, choose an \(N\) such that \(N > \frac{1}{\epsilon^{1/3}}\). Now for all \(n > N\), we have \(n^3 > \frac{1}{\epsilon}\), and hence \(|\frac{1}{n^3}| < \epsilon\). This proves that \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} = 0\) using the definition of the limit of a sequence.