ratio test

Ratio test

Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series.
i) If there exists some $q<1$ such that

$$a_k\ne 0\text{and}\left|\frac{a_{k+1}}{a_k}\right|\le q\text{for almost all }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.
ii) Conversely, if

$$a_k\ne 0\text{and}\left|\frac{a_{k+1}}{a_k}\right|\ge 1\text{for almost all }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is divergent.

The ratio test can be applied in the same situations as the root test, but is sometimes much easier to apply.

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Proof:

i) “For almost all $k$” means the ratio condition holds from some $k_0$ on.
Split off the finite prefix ${\sum }_{k=1}^{k_0-1}{a}_k$; a finite sum cannot change convergence, so w.l.o.g. assume the condition holds from $k=1$.
Rearrange the ratio condition: $|a_{k+1}|\le q|a_k|$.
Each term is at most $q$ times the previous one.
Chaining the one-step bound (induction):

$$\begin{array}{rl}|a_2|& \le q|a_1|\\ |a_3|& \le q|a_2|\le q^2|a_1|\\ & ⋮\\ |a_k|& \le q^{k-1}|a_1|\end{array}$$

So the terms are dominated by a geometric sequence with rate $q$, scaled by $|a_1|$.
By the comparison test against the geometric series:

$$\sum _{k=1}^{\infty }|a_k|\le \sum _{k=1}^{\infty }{q}^{k-1}|a_1|=|a_1|\sum _{k=0}^{\infty }{q}^k$$

Finite for $q\in [0,1)$, so ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.

ii) Rearrange the condition: From some $k_0$ on, $|a_{k+1}|\ge |a_k|$: the absolute values stop decreasing, $(a_k)$ is not a null sequence, hence divergent.

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Equivalently, assuming the limit exists, the ratio test says:

$$a:=\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|$$

i) $a<1⟹$ absolutely convergent
ii) $a>1⟹$ divergent
iii) $a=1⟹$ inconclusive

Note: i) from the definition callout is equivalent to limsup of the ratio $<1$
Note: iii) is inconclusive because if the limit is approached from below, the series converges, whereas if it’s approached from above, the series diverges.

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What is the idea behind the ratio test?

If consecutive terms of the series shrink by a factor of $q<1$ from some point on, then the terms behave like a geometric sequence with ratio $q$, which converges. If they don’t shrink at all (ratio $>1$), then the terms fail to vanish, so the series diverges.

What is the definition of the ratio test?

Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series.
i) If there exists some $q<1$ such that

$$a_k\ne 0\text{and}\left|\frac{a_{k+1}}{a_k}\right|\le q\text{for almost all }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent. Conversely, if that ratio is $\ge 1$ for almost all $k$, the series is divergent.

Equivalently, if $a={\lim }_{k\to \infty }\left|\frac{a_{k+1}}{a_k}\right|$ exists:
$<1⟹$ absolutely convergent,
$>1⟹$ divergent,
$=1⟹$ inconclusive.

Does ${\sum }_{k=1}^{\infty }\frac{2^k}{k!}$ converge or diverge? Show with a convergence test.

Ratio test:

$$\left|\frac{a_{k+1}}{a_k}\right|=\frac{2^{k+1}}{(k+1)!}\cdot \frac{k!}{2^k}=\frac{2}{k+1}\to 0<1$$

so the series converges absolutely.
Same pattern for any $\sum \frac{c^k}{k!}$: the ratio $\frac{c}{k+1}\to 0$, the factorial outgrows every exponential.

Does ${\sum }_{k=1}^{\infty }\frac{k!}{k^k}$ converge or diverge? Show with a convergence test.

That $k^k$ outgrows $k!$ is clear (compare factors), but that only says the terms vanish, which is never sufficient (harmonic series). Convergence needs the rate of the vanishing to be fast enough.
Using the ratio test:

$$\left|\frac{a_{k+1}}{a_k}\right|=\frac{(k+1)!}{(k+1{)}^{k+1}}\cdot \frac{k^k}{k!}=\frac{(k+1)k^k}{(k+1{)}^{k+1}}=\frac{k^k}{(k+1{)}^k}={\left(\frac{k}{(k+1)}\right)}^k={\left(\frac{1}{1+1/k}\right)}^k=\frac{1}{{\left(1+\frac{1}{k}\right)}^k}\to \frac{1}{e}<1$$

So the series converges absolutely: $k^k$ beats $k!$, by a factor of about $e^k$.