Ratio Test

Theorem

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

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

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

Proof

From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$ otherwise $\left\vert{\dfrac {a_{n+1}}{a_n}}\right\vert$ is not defined.

• Suppose $l < 1 $.

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

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

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{a_n}\right\vert\)	\(=\)	\(\displaystyle \left\vert{\frac {a_n}{a_{n-1} } }\right\vert \left\vert{\frac {a_{n-1} }{a_{n-2} } }\right\vert \cdots \left\vert{\frac {a_{N+2} }{a_{N+1} } }\right\vert \left\vert{a_{N+1} }\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \left({l + \epsilon}\right)^{n - N - 1} \left\vert{a_{N+1} }\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

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

$\blacksquare$

• Suppose $l > 1$.

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{a_n}\right\vert\)	\(=\)	\(\displaystyle \left\vert{\frac {a_n}{a_{n-1} } }\right\vert \left\vert{\frac {a_{n-1} }{a_{n-2} } }\right\vert \cdots \left\vert{\frac {a_{N+2} }{a_{N+1} } }\right\vert \left\vert{a_{N+1} }\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(>\)	\(\displaystyle \left({l - \epsilon}\right)^{n - N + 1} \left\vert{a_{N+1} }\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

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

$\blacksquare$

Notes

If $l = 1$, it is impossible to say, without further analysis, whether $\displaystyle \sum_{n=1}^\infty a_n$ converges or not.

If $\displaystyle \left\vert{\frac {a_{n+1}}{a_n}}\right\vert \to \infty$ as $n \to \infty$, then of course $\displaystyle \sum_{n=1}^\infty a_n$ diverges.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 6.17$