Ratio Test

Theorem
Let $$\sum_{n=1}^\infty a_n$$ be a series in $\mathbb{R}$.

Let the sequence $$\left \langle {a_n} \right \rangle$$ satisfy $$\lim_{n \to \infty} \left|{\frac {a_{n+1}}{a_n}}\right| = l$$.


 * If $$l > 1 $$, then $$\sum_{n=1}^\infty a_n$$ diverges.
 * If $$l < 1 $$, then $$\sum_{n=1}^\infty a_n$$ converges.

Proof

 * Suppose $$l < 1 $$.

Let us take $$\epsilon > 0$$ such that $$l + \epsilon < 1$$.

Then $$\exists N: \forall n > N: \left|{\frac {a_n}{a_{n-1}}}\right| < l + \epsilon$$.

Thus:

$$ $$

By Sum of Powers, $$\sum_{n=1}^\infty \left({l + \epsilon}\right)^n$$ converges.

So by the the corollary to the comparison test, it follows that $$\sum_{n=1}^\infty a_n$$ converges too.


 * Suppose $$l > 1$$.

Let us take $$\epsilon > 0$$ small enough that $$l - \epsilon > 1$$.

Then, for a sufficiently large $$N$$, we have:

$$ $$

But $$\left({l - \epsilon}\right)^{n - N + 1} \left|{a_{N+1}}\right| \to \infty$$ as $$n \to \infty$$.

So $$\sum_{n=1}^\infty a_n$$ diverges.