Editing Ratio test (section)

== Examples ==
=== Convergent because ''L'' < 1 ===
Consider the series

:<math>\sum_{n=1}^\infty\frac{n}{e^n}</math>

Applying the ratio test, one computes the limit

:<math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right| = \frac{1}{e} < 1.</math>

Since this limit is less than 1, the series converges.

=== Divergent because ''L'' > 1 ===
Consider the series

:<math>\sum_{n=1}^\infty\frac{e^n}{n}.</math>

Putting this into the ratio test:

:<math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}} \right|
= e > 1.</math>

Thus the series diverges.

=== Inconclusive because ''L'' = 1 ===
Consider the three series

:<math>\sum_{n=1}^\infty 1,</math>
:<math>\sum_{n=1}^\infty \frac{1}{n^2},</math>
:<math>\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}.</math>

The first series ([[1 + 1 + 1 + 1 + ⋯]]) diverges, the second (the one central to the [[Basel problem]]) converges absolutely and the third (the [[alternating harmonic series]]) converges conditionally. However, the term-by-term magnitude ratios <math>\left|\frac{a_{n+1}}{a_n}\right|</math> of the three series are <math>1,</math> &nbsp;&nbsp;<math>\frac{n^2}{(n+1)^2}</math> &nbsp;&nbsp; and &nbsp;&nbsp;<math>\frac{n}{n+1}</math>. So, in all three, the limit <math>\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|</math> is equal to 1. This illustrates that when ''L'' = 1, the series may converge or diverge: the ratio test is inconclusive.  In such cases, more refined tests are required to determine convergence or divergence.