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 formula for the terms in sequence whose sum we are trying to find. We can observe something interesting when $k$ is large by looking at the ratio $\left |\frac {a_{n+1}}{a_n}\right |$.

Notice that by treating the division by $a_k$ as multiplication by its reciprocal,

\[ \lim _{k \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \lim _{k \to \infty } \frac {k+1}{2^{k+1}} \cdot \frac {2^k}{k}. \]

After rearranging a little, we have

\[ \lim _{k \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \lim _{k \to \infty } \frac {k+1}{k} \cdot \frac {2^k}{2^{k+1}}, \]

and we can now finish computing the limit.

\[ \lim _{k \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \lim _{k \to \infty } \frac {k+1}{k} \cdot \frac {\cancel {2^k}}{\cancel {2^k} \cdot 2^1} = \answer [given]{\frac {1}{2}}. \]

Thus, when $k$ is large, $a_{k+1}$ is pretty close to halfdouble of $a_{k}$. So, when we choose a very large whole number $N$, $a_{N+1}$ should be approximately $\frac {1}{2} a_N$, and this should occur for all successive terms in the sequence $\{a_k\}$. That is,

\[ a_{k+1} \approx \frac {1}{2} \cdot a_k \]

for all $k>N$.

We can then observe the following. \begin{align*} a_{N+1} &\approx \frac {1}{2} \cdot a_N \\ a_{N+2} &\approx \frac {1}{2} \cdot a_{N+1} \approx \frac {1}{2} \cdot \left (\frac {1}{2} \cdot a_N\right ) = \answer [given]{\left (\frac {1}{2}\right )^2} \cdot a_N \\ a_{N+3} &\approx \frac {1}{2} \cdot a_{N+2} \approx \frac {1}{2} \cdot \left ( \left (\frac {1}{2}\right )^2 \cdot a_N\right ) = \answer [given]{\left (\frac {1}{2}\right )^3} \cdot a_N \\ \vdots \end{align*}

We thus have that

\[ \sum _{k=N}^{\infty } \frac {k}{2^k} \approx \sum _{k=N}^{\infty } \left (\frac {1}{2}\right )^k \cdot a_N. \]

In words, the tail of the sequence $\{a_k\}_{k=N}$ 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

As it turns out, the above ideas can be formalized and written as a theorem. The proof of this theorem is slightly beyond the scope of the course, but really only involves introducing formal mathematical language to make the above observations precise.

The Ratio Test Consider $\sum _{k=k_0}^\infty a_k$ and set

\[ L = \lim _{n \to \infty } \left | \frac {a_{n+1}}{a_n} \right |. \]

If $0 \leq L < 1$, then the series converges.
If $L>1$ or $L$ is infinite, then the series diverges.
If $L = 1$, the test is inconclusive; 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 \neq 1 $, 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, which we will demonstrate with specific examples in a bit.

Now that you have the basic idea, we give examples showing the following.

Convergence by the ratio test.
Divergence by the ratio test.
A divergent series for which the ratio test is inconclusive.
A convergent series for which the ratio test is inconclusive.

In these examples, pay attention to how the ratio of different types of terms behave and simplify.

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

We will attempt to use the ratio test. Setting $a_n = \frac {2^n}{n!}$, we compute \begin{align*} \lim _{n \to \infty } \frac {a_{n+1}}{a_n} &= \lim _{n \to \infty } \frac {2^{n+1}}{(n+1)!} \cdot \frac {n!}{2^n} \\ &= \lim _{n \to \infty } \frac {2^{n+1}}{2^n} \cdot \frac {n!}{(n+1)!} \\ &= \lim _{n \to \infty } \frac {\cancel {2^n}\cdot 2^1}{\cancel {2^n}} \cdot \frac {\cancel {n!}}{(n+1) \cdot \cancel {n!} } \\ &= \answer [given]{0} \end{align*}

So, the ratio test guarantees that $ \sum _{k=4}^\infty \frac {2^k}{k!}$ is convergent guarantees that $\sum _{k=4}^\infty \frac {2^k}{k!}$ 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.

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 {5^{1+k}}{k \cdot 2^{2k+1}}$. Discuss the convergence of this series.

We will attempt to use the ratio test. Setting $a_n = \frac {5^{1+n}}{n \cdot 2^{2n+1}}$, we compute \begin{align*} \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | &= \lim _{n \to \infty } \frac {5^{n+2}}{(n+1)2^{2(n+1)+1}} \cdot \frac {n 2^{2n+1}}{5^{n+1}}\\ &=\lim _{n \to \infty } \frac {5^{n+2}}{5^{n+1}} \cdot \frac {2^{2n+1}}{2^{2n+3}} \cdot \frac {n}{n+1}\\ &=\lim _{n \to \infty } \frac {\cancel {5^{n+1}}\cdot 5^1}{\cancel {5^{n+1}}} \cdot \frac {\cancel {2^{2n+1}}}{\cancel {2^{2n+1}}\cdot 2^2} \cdot \frac {n}{n+1}\\ &=\lim _{n \to \infty } \frac {5}{4} \cdot \frac {n}{n+1}\\ &=\frac {5}{4} \end{align*}

So, the ratio test guarantees that $ \sum _{k=1}^\infty \frac {5^{1+k}}{k \cdot 2^{2k+1}}$ is convergent guarantees that $ \sum _{k=1}^\infty \frac {5^{1+k}}{k \cdot 2^{2k+1}}$ 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.

We now turn to two examples where the ratio test will be inconclusive.

Consider $\sum _{k=1}^\infty k^{1000}$. Discuss the convergence of this series.

Note that this question is slightly ridiculous; since $\lim _{n \to \infty } n^{1000} = \infty $, the series will diverge by the divergence test. However, if we try to use the ratio test, what happens?

\[ L = \lim _{n \to \infty } \frac {a_{n+1}}{a_n} = \lim _{n \to \infty } \frac {(n+1)^{1000}}{n^{1000}} = \lim _{n \to \infty } \left (\frac {n+1}{n}\right )^{1000} = \answer [given]{1} \]

So, the ratio test guarantees $\sum _{k=1}^\infty k^{1000}$ is convergent guarantees $\sum _{k=1}^\infty k^{1000}$ is divergent is inconclusive.

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

We will attempt to use the ratio test. Setting $a_n = \frac {1}{n^2+n}$, notice \begin{align*} \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | &= \lim _{n \to \infty } \frac {1}{(n+1)^2+(n+1)} \cdot \frac {n^2+n}{1} \\ &= \lim _{n \to \infty } \frac {n^2+n}{(n+1)^2+(n+1)} \end{align*}

We could do some algebra here, but notice that the dominant term in both the numerator and denominator will be $n^2$. As such, we can conclude that the limit is $1$ without further computation.

The ratio test guarantees $\sum _{k=1}^\infty \frac {1}{k^2+k}$ is convergent guarantees $\sum _{k=1}^\infty \frac {1}{k^2+k}$ is divergent is inconclusive.

In order to determine what happens here, we can actually use partial fraction decomposition to show that

\[ \frac {1}{k^2+k} = \frac {1}{k}-\frac {1}{k+1}. \]

As such, we can write

\[\sum _{k=1}^{\infty } \frac {1}{k^2+k} = \sum _{k=1}^{\infty } \left (\frac {1}{k}-\frac {1}{k+1}\right ),\]

and recognize this as a telescoping series. By setting $s_n = \sum _{k=1}^n$ and writing out several terms (as done in an earlier section), we can find an explicit formula

\[ s_n = 1 - \frac {1}{n+1}. \]

Since $\lim _{n \to \infty } s_n = 1$, we conclude that the series $\sum _{k=1}^{\infty } \frac {1}{k^2+k}$ converges to 1.

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

In the previous examples, we studied sequences $\{a_n\}$ whose terms involved products of factorials, exponentials, and polynomials. One interesting observation to note is that the ratio of polynomials in the examples above did not affect the value of $\lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right |$ and this was perhaps most readily evident in the wildly divergent series $\sum _{k=1}^{\infty } k^{1000}$.

As it turns out, this fact will be important in an observation we make later, so we dignify it with a theorem here.

Suppose that $p(x)$ is a polynomial. Then,

\[ \lim _{n \to \infty } \frac {p(n+1)}{p(n)} = 1. \]

This theorem guarantees that if $\{a_n\}$ is a sequence whose terms involve products or quotients of polynomials, these polynomial terms will have no effect on the limit that is computed using the ratio test.

Let $a_n = \frac {n^3}{3^n}$ and $b_n = \frac {1}{3^n}$. According to the theorem, $\lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right |$ and $ \lim _{n \to \infty } \left |\frac {b_{n+1}}{b_n}\right |$ should be equal. To verify this, notice that

\[ \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \lim _{n \to \infty } \frac {(n+1)^3}{3^{n+1}} \cdot \frac {3^n}{n^3} = \lim _{n \to \infty } \frac {\cancel {3^n}}{\cancel {3^n} \cdot 3} \cdot \left (\frac {n+1}{n}\right )^3 = \frac {1}{3} \]

and also that

\[ \lim _{n \to \infty } \left |\frac {b_{n+1}}{b_n}\right | = \lim _{n \to \infty } \frac {1}{3^{n+1}} \cdot \frac {3^n}{1} = \lim _{n \to \infty } \frac {3^n}{3^{n+1}} = \lim _{n \to \infty } \frac {\cancel {3^n}}{\cancel {3^n} \cdot 3} = \frac {1}{3} \]

This example also motivates an important fact. Notice that for each term with $n >1$, we have $b_n < a_n$, but the limits necessary to compute for the ratio test are equal. As such, we observe the following.

The ratio test can only tell us that a series converges; it cannot give the value to which the series converges.

The theorem also leads to a very important observation about when we should attempt to use the ratio test. A series must have a term that grows at least exponentially in order for the Ratio Test to have a chance to be conclusive!

More explicitly, let $p, q > 0$, $a>1$, and consider the growth rates results for sequences:

Without performing any calculations, determine which of the following series would the Ratio Test would be conclusive. That is, which of the following would either converge or diverge as a consequence of the Ratio Test?

$\sum _{k=1}^{\infty } \frac {\sqrt {k^2+4}}{k^5+2\sqrt {k}+1}$ $\sum _{k=1}^{\infty } \frac {\ln (k) +k^2}{k!}$ $\sum _{k=1}^{\infty } \frac {k^{10}}{5^k}$ $\sum _{k=1}^{\infty } \frac {(\ln k)^{80}}{k^5}$ $\sum _{k=1}^{\infty } \frac {2^k}{k^k}$

1 Summary

We summarize the important results about the ratio test. Consider a series $\sum _{k=k_0}^{\infty } a_k$.

If $p(x)$ is a polynomial, $\lim _{n \to \infty } \frac {p(n+1)}{p(n)} = 1$, so the ratio test will only be conclusive if $a_k$ has a factor that grows at least exponentially (according to the growth rates results).
To use the test, set $L = \lim _{n \to \infty } \left | \frac {a_{n+1}}{a_n} \right |.$
- If $0 \leq L < 1$, then the series converges.
- If $L>1$ or $L$ is infinite, then the series diverges.
- If $L = 1$, the test is inconclusive; the series could diverge or converge.
In the case where a series converges, the ratio test gives no information about the value of the series.

2025-01-06 20:24:40