Question

Suppose that \(\sum a_{n}\) and \(\sum b_{n}\) are series with positive terms. Prove that if \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\) and \(\sum b_{n}\) converges, \(\Sigma a_{n}\) also converges.

Short answer

Since \(a_n\) is ultimately less or equal a certain fraction of \(b_n\) and the series \(\sum b_n\) converges, then \(\sum a_n\) also converges. Therefore, given \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\) and \(\sum b_{n}\) converges, \(\sum a_{n}\) must also converge.

Step-by-step solution

Assert the Limit Condition

Given we know that \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\). This essentially means that for each \(\epsilon > 0\) there exists an \(N \in \mathbb{N}\) such that if \(n > N\) then \(|\frac{a_{n}}{b_{n}}| \leq \epsilon \). Instead of \(\epsilon\) we can use a positive number \(l\) satisfying \(0 < l < 1\). Hence, for all \(n > N\), \(\frac{a_{n}}{b_{n}} \leq l\) which implies that \(a_n \leq l * b_n\) for all \(n > N\).

Apply Calculation

What \(a_n \leq l * b_n\) means that, for all \(n > N\), the \(n-th\) term of the \(a\) series is less than or equal to a fraction of corresponding \(n-th\) term of the \(b\) series. Given \(\sum b_n\) converges, it implies, by the comparison test that \(0 \leq a_n \leq l * b_n\) for all \(n > N\) also converges.

Conclude The Proof

In conclusion, as per the conditions given in the problem, if \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\) and \(\sum b_{n}\) converges, then \(\sum a_{n}\) must converge. This is because we determined that \(a_n\) is ultimately less or equal a fraction \(l\) of \(b_n\), and since the series \(\sum b_n\) converges, then \(\sum a_n\) also converges by comparison test.