# College Math Teaching

## April 3, 2011

### Infinite Series: the Root and Ratio tests

Filed under: advanced mathematics, analysis, calculus, infinite series, series — collegemathteaching @ 3:29 am

This post is written mostly for those who are new to teaching calculus rather than for students learning calculus for the first time. Experienced calculus teachers and those whose analysis background is still fresh will likely be bored. 🙂

The setting will be series $\Sigma^\infty_{k=1} a_k$ with $a_k > 0$ We will use the usual notion of convergence; that is, the series converges if the sequence of partial sums converge. If that last statement puzzles you, there are other non-standard notions of convergence.

I’ll give the usual statements of the root test and the ratio test, given $\Sigma^\infty_{k=1} a_k$ with $a_k > 0$

Root Test
Suppose $lim_{k \rightarrow \infty}(a_k)^{\frac{1}{k}} = c$. If $c >1$ the series diverges. If $c < 1$ the series converges. If $c = 1$ the test is inconclusive.

Ratio Test
Suppose $lim_{k \rightarrow \infty} a_{k+1}/a_k = c$. If $c >1$ the series diverges. If $c < 1$ the series converges. If $c = 1$ the test is inconclusive.

Quick examples of how these tests are used
Example one: show that $\Sigma^{\infty}_{k=1} (x/k)^{k}$ converges for all $x$. Apply the root test and note $lim_{k \rightarrow \infty} ((x/k))^{k})^{1/k} = lim_{k \rightarrow \infty} ((x/k) = 0$ for all $x \geq 0$ hence the series converges absolutely for all $x$.

Example two: show that show that $\Sigma^{\infty}_{k=1} (x^k/k!)$ converges for all $x$. Consider $lim_{k \rightarrow \infty} (x^{k+1}/(k+1)!)/((x^k)/k!) = lim_{k \rightarrow \infty} x/(k+1) = 0 < 1$ for all $x \geq 0$ hence the series converges absolutely for all $x$.

However these tests, as taught, are often more limited than they need to be. For example, if one considers the series $\Sigma^{\infty}_{k=1} (|sin(k)|/2)^{k}$ the root test, as stated, doesn't apply as $lim_{k \rightarrow \infty} ((|sin(k)|)/2)^{k})^{1/k}$ fails to exist, though it is clear that the $sup{((|sin(k)|)/2)^{k})^{1/k}} = 1/2$ and the series is dominated by the convergent geometric series $\Sigma^{\infty}_{k=1} (1/2)^k$ and therefore converges.

There is also a common misconception that the root test and the ratio tests are equivalent. They aren’t; in fact, we’ll show that if a series passes the ratio test, it will also pass the root test but that the reverse is false. We’ll also provide an easy to understand stronger version of these two tests. The stronger versions should be easily within the grasp of the best calculus students and within the grasp of beginning analysis/advanced calculus students.

Note: there is nothing original here; I can recommend the books Counterexamples in Analysis by Bernard R. Gelbaum and Theory and Application of Infinite Series by Konrad Knop for calculus instructors who aren’t analysts. Much of what I say can be found there in one form or another.

The proofs of the tests and what can be learned
Basically, both proofs are merely basic comparisons with a convergent geometric series.

Proof of the root test (convergence)
If $lim_{k \rightarrow \infty} (a_k)^{1/k} = c$ with $c < 1$ then there exists some $d <1$ and some index $N$ such that for all $n > N$, ${a_n}^{1/n} < d$. Hence for all $n> N$, ${a_n} < d^n$ where $d < 1$. Therefore the series converges by a direct comparison with the convergent geometric series $\Sigma^{\infty}_{k = N} d^{k}$.

Note the following: requiring $lim_{k \rightarrow \infty} (a_k)^{1/k}$ to exist is overkill; what is important that, for some index $N$, $k$, $sup_{k>N} (a_k)^{1/k} < c < 1$. This is enough for the comparison with the geometric series to work. In fact, in the language of analysis, we can replace the limit condition with $limsup (a_k)^{1/k} = c < 1$. If you are fuzzy about limit superior, this is a good reference.

In fact, we can weaken the hypothesis a bit further. Since the convergence of a series really depends on the convergence of the “tail” of a series, we really only need: for some index $N$ and all $n > N$, $limsup (a_{N+k})^{1/k} = c < 1$, $k \in \{0, 1, 2,...\}$. This point may seem pedantic but we’ll use this in just a bit.

Note: we haven’t talked about divergence; one can work out a stronger test for divergence by using limit inferior.

Proof of the ratio test (convergence)
We’ll prove the stronger version of the ratio test: if $limsup a_{k+1}/a_k = c < 1$ then there is an index $N$ and some number $d < 1$ such that for all $n\geq N$, $a_{n+1}/a_n < d$.
Simple algebra implies that $a_{n+1} < (a_n)d$ and $a_{n+2} < (a_{n+1})d < (a_n)d^2$ and in general $a_{n+j} < (a_n)d^j$. Hence the series $\Sigma ^{\infty}_{k = N} a_k$ is dominated by $(a_N)(\Sigma ^{\infty}_{k = 0} d^k)$ which is a convergent geometric series.

Comparing the root and the ratio tests
Consider the convergent series $1/2 + (1/3)^2 + (1/2)^3 + (1/3)^4 +.....+ (1/2)^{2k- 1} + (1/3)^{2k}....$.
Then clearly $limsup a_{k} = 1/2$ hence the root test works. But the ratio test yields the following:
$(a_{2k+1})/a_{2k} = (1/2)^{2k+1}/(1/3)^{2k} = (3/2)^{2k}/2$ which tends to infinity as $k$ goes to infinity. Note: since the limit does not exist, the traditional ratio test doesn’t apply. The limit inferior tends to zero so a strengthened ratio test doesn’t imply divergence.

So the root test is not equivalent to the ratio test.

But suppose the ratio test yields convergence; that is:
$limsup a_{k+1}/a_k = c < 1$. Then by the same arguments used in the proof:
$a_{N+j} < (a_N)d^j$. Then we can take $j'th$ roots of both sides and note: $(a_{n+j})^{1/j} < d(a_n)^{1/j} < d$ hence the weakened hypothesis of the root test is met.

That is, the root test is a stronger test than the ratio test, though, of course, it is sometimes more difficult to apply.

We’ll state the tests in the stronger form for convergence; the base assumption that $\Sigma a_{k}$ has positive terms:

Root test: if there exists an index $N$ such that for all $n \leq N$ we have $limsup_{j} (a_{N+j})^{1/j} \leq c < 1$ then the series converges.

Ratio test: if $limsup (a_{k+1})/a_{k} \leq c < 1$ then the series converges.

It is a routine exercise to restate these tests in stronger form for divergence.