Prove the ratio test.

Steps to Solve

1. Define the series and limit

Consider an infinite series of the form $$\sum_{n=1}^{\infty} a_n$$ where (a_n) is a sequence of real or complex numbers. We will apply the ratio test by defining the limit:

$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

2. Evaluate different cases of the limit

Now we will analyze the limit (L) to determine the behavior of the series.

• If $L < 1$, the series converges absolutely.

• If $L > 1$ (or $L = \infty$), the series diverges.

• If $L = 1$, the test is inconclusive, and we cannot determine convergence or divergence from this test alone.

3. Proof of absolute convergence for (L < 1)

For $L < 1$, there exists a positive number (r) such that (L < r < 1).

This implies:

$$ \left| \frac{a_{n+1}}{a_n} \right| < r $$

for sufficiently large (n). By rearranging:

$$ |a_{n+1}| < r |a_n| $$

This suggests that the terms of the series decrease in size sufficiently quickly and aids in showing that the series converges absolutely through comparison with a convergent geometric series.

4. Proof of divergence for (L > 1)

For $L > 1$, there exists (R > 1) such that

$$ \left| \frac{a_{n+1}}{a_n} \right| > R $$

for sufficiently large (n).

This means:

$$ |a_{n+1}| > R |a_n| $$

where subsequently, the terms of the series cannot get smaller to approach 0, indicating that the series diverges.

5. Concluding the proof

So, we conclude that the ratio test shows that a series $$\sum_{n=1}^{\infty} a_n$$ converges absolutely if (L < 1) and diverges if (L > 1). The test is inconclusive if (L = 1).