Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(a_{n} \leq b_{n}+c_{n}\) and \(\sum_{n=1}^{\infty} a_{n}\) diverges, then the series \(\sum_{n=1}^{\infty} b_{n}\) and \(\sum_{n=1}^{\infty} c_{n}\) both diverge. (Assume that the terms of all three series are positive.)

Short answer

The statement is false. Convergence or divergence of \(b_{n}\) and \(c_{n}\) can't be determined based on the divergence of \(a_{n}\) and the inequality \(a_{n} \leq b_{n}+c_{n}\) alone. A counterexample is \(a_{n} = 1/n\), \(b_{n} = 1/(n^2)\), and \(c_{n} = 1/(n^3)\) where the conditions hold true, but \(b_{n}\) and \(c_{n}\) both converge.

Step-by-step solution

Initialize series

Let \(a_{n} = 1/n\), \(b_{n} = 1/(n^2)\), and \(c_{n} = 1/(n^3)\). These series are all positive as required.

Check First Condition

We need to check if \(a_{n} \leq b_{n}+c_{n}\). Given our chosen series, \(1/n \leq 1/(n^2)+1/(n^3)\). This is true because when simplified, it is equivalent to saying \(1 \leq 1\), which is an equality.

Check the Divergence of \(a_{n}\)

We must determine if \(\sum_{n=1}^{\infty} 1/n\), also known as the Harmonic Series, diverges. The Harmonic Series is well-known to be a divergent series.

Test the Convergence/Divergence of \(b_{n}\) and \(c_{n}\)

We know that \(\sum_{n=1}^{\infty} 1/n^2\) converges based on the p-series test, where p > 1. Similarly, \(\sum_{n=1}^{\infty} 1/n^3\) also converges by the p-series test.

Conclude

Given that \(a_{n}\) is less than or equal to the sum of \(b_{n}\) and \(c_{n}\), and \(a_{n}\) diverges, yet both \(b_{n}\) and \(c_{n}\) converge, the statement is false. It's not necessarily true that if \(a_{n} \leq b_{n}+c_{n}\) and \(a_{n}\) diverges, then the series \(\sum_{n=1}^{\infty} b_{n}\) and \(\sum_{n=1}^{\infty} c_{n}\) both diverge.