Prove the ratio test. Hint : If \(\left|a_{n+1} / a_{n}\right| \rightarrow \rho<1\), take \(\sigma\) so that \(\rho<\sigma<1\). Then \(\left|a_{n+1} / a_{n}\right|<\sigma\) if \(n\) is largc, say \(n \geq N\). This means that we have \(\left|a_{y-1}\right|<\sigma\left|a_{N}\right|\), \(\left|a_{N+2}\right|1\) diverges. Hint: Take \(\rho>\sigma>1\), and use the prcliminary test.

A series \(\ sum_{n=1}^{\infty} a_n\) converges if the sum of its terms tends to a finite number as the number of terms increases. To determine whether a series converges, we often use tests like the ratio test. Essentially, if the terms of a series become smaller quickly enough, the series will add up to a finite value. \(\textbf{Ratio Test for Convergence:}\)Consider a series \(\ sum_{n=1}^{\infty} a_n\). The ratio test involves evaluating the limit: \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \rho\). Based on \(\rho\):\

• If \(\rho < 1\), the series converges.
• If \(\rho > 1\), the series diverges.
• If \(\rho = 1\), the test is inconclusive.

When \(\rho < 1\), there exists a value \(\sigma\) such that \(\rho < \sigma < 1\), making the terms \(\left| \frac{a_{n+1}}{a_n} \right| < \sigma\) for large enough n.

A geometric series has the form:\(\ sum_{n=0}^{\infty} ar^n \), where a is the first term and r is the common ratio. The sum of a geometric series is finite if the absolute value of r is less than 1, and it is given by the formula:\(\ sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \).
When we use the ratio test to determine the convergence of a series, we often compare it to a geometric series with ratio \(\sigma < 1\). If the terms of our series are less than those of a convergent geometric series for all sufficiently large n, our series must also converge. This comparison helps to visualize why the series will sum to a finite value.
Understanding geometric series makes it easier to understand why the ratio test works and provides a practical way to assess a series's behavior.

A series \(\ sum_{n=1}^{\infty} a_n\) diverges if the sum does not tend to a finite value. This can happen if the terms do not get smaller quickly enough as n increases. \(\textbf{Ratio Test for Divergence:}\)When applying the ratio test, if \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \rho > 1\), the series must diverge because the terms will eventually get larger, causing the sum to increase indefinitely.

• If \(\left| \frac{a_{n+1}}{a_n} \right|\) consistently exceeds a value greater than 1, the terms will grow without bound.
• If the terms of the series do not approach zero, the series cannot add up to a finite value.

For example, with \(\rho > 1\), we can choose \(\sigma\) such that \(\sigma > 1\), leading to \(\left| a_{n+1} \right| > \sigma \left| a_{n} \right|\), and thus the terms grow larger.

The limit comparison test is a method for determining the convergence or divergence of a series by comparing it to a known benchmark series. To apply this test:

• Consider two series \(\ sum_{n=1}^{\infty} a_n\) and \(\ sum_{n=1}^{\infty} b_n\).
• Calculate \(\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = L\).

Based on the value of L:

• If \(L > 0\) and L is finite (a positive real number), then both series either converge or diverge together.
• If \(L = 0\) or \(L = \infty \), then the test may be inconclusive.

The limit comparison test is particularly helpful when direct application of other convergence tests is challenging. It simplifies complex series by comparing them with simpler, well-understood benchmark series, facilitating easier conclusions about their behavior.