The ratio test

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Some infinite series can be compared to geometric series.

As mathematicians, we are explorers. We explore the implications of seemingly simple quantitative facts.

Consider the infinite series $\sum _{k=0}^\infty \frac {k}{2^k}$ Let $a_k = \answer [given]{\frac {k}{2^k}}$ be the sequence of terms of this series. When $k$ is large, $a_{k+1}$ is pretty close to halfdouble of $a_{k}$ . The effect of the numerator increasing by $1$ is dwarfed by the effect of the denominator being doubled. We can formalize this by looking at the ratio of consecutive terms: $\lim _{k \to \infty } \frac {a_{k+1}}{a_k} = \answer [given]{\frac {1}{2}}$ When we choose a very large whole number $N$ , $a_{k+1} \approx a_k\cdot \frac {1}{2}$ for $k>N$ , and so we get the following approximations: $\begin{align*} a_{N+1} &\approx a_N\cdot \frac{1}{2} \\ a_{N+2} &\approx a_N\cdot \answer[given]{\frac{1}{2^2}} \\ a_{N+3} &\approx a_N\cdot \answer[given]{\frac{1}{2^3}}\\ \vdots \end{align*}$ In other words, the tail of the sequence $(a_k)$ beginning with $k=N$ is “approximately” a geometric series with ratio $\frac {1}{2}$ .

Does a geometric series with ratio $\frac {1}{2}$ converge or diverge?

converge diverge

Given your answer above, do you suspect that the original sum $\sum _{k=0}^\infty \frac {k}{2^k}$ converges or diverges?

converge diverge

The above exploration motivates the following theorem. The proof of this theorem is slightly beyond the scope of the course.

The Ratio Test Let $\sum _{k=0}^\infty a_k$ be an infinite series with positive terms. If $r = \lim _{k \to \infty } \frac {a_{k+1}}{a_k}$ exists, then we can conclude the following.

If $0 \leq r < 1$ , then the series converges.
If $r>1$ , then the series diverges.
If $r = 1$ , then we learn nothing: the series could diverge or converge.

Note that this is easy to remember if you just use the following heuristic.

If the ratio test gives a limit of $r$ , then the series is like a geometric series of ratio $r$ .

The case of $r=1$ is an “edge” case, and can go either way. Now that you have the basic idea, we give examples showing:

The ratio test indicating convergence.
The ratio test indicating divergence.
The ratio test being inconclusive, but the series actually converges.
The ratio test being inconclusive, but the series actually diverges.

It is important that examples illustrating the final two behaviors exist, because it shows that the ratio test really is inconclusive in the case $r=1$ .

Consider: $\sum _{k=4}^\infty \frac {2^k}{k!}$ Discuss the convergence of this series.

We will attempt to use the ratio test. Setting $a_k = \frac {2^k}{k!}$ . Write with me. $\lim _{k \to \infty } \frac {a_{k+1}}{a_k} = \answer [given]{0}$ So, the ratio test says the series is convergent says the series is divergent gives no information in this case, but we know the series is convergent through some other method gives no information in this case, but we know the series is divergent through some other method .

$\begin{align*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} &= \lim_{k \to \infty} \frac{2^{k+1}}{(k+1)!} \frac{k!}{2^k}\\ &=\lim_{k \to \infty} \frac{2}{k+1}\\ &=0 \end{align*}$ So the series is convergent by the ratio test. Note that this shows that $k!$ grows much faster than the exponential function $2^k$ .

Consider: $\sum _{k=1}^\infty \frac {1}{k}$ Discuss the convergence of this series.

We will attempt to use the ratio test. Setting $a_k = \frac {1}{k}$ . Write with me. $\lim _{k \to \infty } \frac {a_{k+1}}{a_k} = \answer [given]{1}$ So, the ratio test says the series is convergent says the series is divergent gives no information in this case, but we know the series is convergent using the integral test gives no information in this case, but we know the series is divergent using the integral test .

$\begin{align*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} &= \lim_{k \to \infty} \frac{k+1}{k}\\ &=1 \end{align*}$ So the ratio test gives no information. However, we know that the harmonic series is divergent (we proved this using the integral test).

Consider: $\sum _{k=1}^\infty \frac {5^{1+k}}{k 2^{2k+1}}$ Discuss the convergence of this series.

We will attempt to use the ratio test. Setting $a_k = \frac {5^{1+k}}{k 2^{2k+1}}$ . Write with me. $\lim _{k \to \infty } \frac {a_{k+1}}{a_k} = \answer [given]{\frac {5}{4}}$ So, the ratio test says the series is convergent says the series is divergent gives no information in this case, but we know the series is convergent through some other method gives no information in this case, but we know the series is divergent through some other method .

$\begin{align*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} &= \lim_{k \to \infty} \frac{5^{k+2}}{(k+1)2^{2(k+1)+1}} \frac{k 2^{2k+1}}{5^{1+k}}\\ &=\lim_{k \to \infty} \frac{5^{k+2}}{5^{1+k}} \frac{2^{2k+1}}{2^{2k+3}} \frac{k}{k+1}\\ &=\lim_{k \to \infty} \frac{5}{4} \frac{k}{k+1}\\ &=\frac{5}{4} \end{align*}$ So the series is divergent by the ratio test.

Consider: $\sum _{k=1}^\infty \frac {1}{k^2}$ Discuss the convergence of this series.

We will attempt to use the ratio test. Setting $a_k = \frac {1}{k^2}$ . Write with me. $\lim _{k \to \infty } \frac {a_{k+1}}{a_k} = \answer [given]{1}$ So, the ratio test says the series is convergent says the series is divergent gives no information in this case, but we know the series is convergent through some other method gives no information in this case, but we know the series is divergent through some other method .

$\begin{align*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} &= \lim_{k \to \infty} \frac{(k+1)^2}{k^2}\\ &= \lim_{k \to \infty} \left( \frac{k+1}{k}\right)^2\\ &= \lim_{k \to \infty} \left( 1+\frac{1}{k}\right)^2\\ &= 1 \end{align*}$ So the ratio test gives no information. However, we know that this series is convergent by the $p$ series test with $p=2$ (which ultimately derives from the integral test).