Ознака Д'Аламбера

Like the root test, the ratio test gives us another method of seeing when a series converges or diverges. And the reason why we introduce this after the root test is that we can show (and we will at the end of this section) that the root test can be used to prove the ratio test.

But wait, why do we need another test, if the ratio test follows from the root test? The next example gives us a clue:

If we try to solve this question through the root test, we have $$L=\lim_{n\to\infty} \left(\frac{1}{n!}\right)^{\frac{1}{n}}=\lim_{n\to\infty} \frac{1}{(n!)^{\frac{1}{n}}}=\frac{1}{\lim_{n\to\infty} (n!)^{\frac{1}{n}}}.$$ But what exactly is that denominator? One way to find out is by using Stirling's approximation, which states that for large values of $n$, $n!$ is very close to $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$. But that's pretty complicated! Is there another way to see whether this series converges?

Thankfully, the answer is yes, and it's called the ratio test. The test states the following:

Theorem For the series $\sum a_n$, consider the number $$L=\lim_{n\to\infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|.$$ Then we have the following:

1. If $L<1$, then $\sum a_n$ converges absolutely.

2. If $L>1$, the series diverges.

3. If $L=1$, then the test is inconclusive.

Knowing this theorem, let's return to the example problem above, and see whether the series $$\sum_{n=1}^{\infty} \frac{1}{n!}$$ converges or diverges. Since $$(n+1)!=(n+1)\cdot [n\cdot (n-1)\cdots 2\cdot 1]=(n+1)\cdot n!,$$ we have $$L=\lim_{n\to\infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=\lim_{n\to\infty} \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=\lim_{n\to\infty} \frac{\frac{1}{(n+1)\cdot n!}}{\frac{1}{n!}}=\lim_{n\to\infty} \frac{1}{n+1}=0.$$ So by the ratio test, the series converges absolutely.

Wow, that was a lot nicer and easier! Like this case, there are many instances in which the ratio test is more fitting to use than the root test. Usually, whenever a series involves factorials or exponential functions, the ratio test is a good idea to start with.

Now you try!

Let's work this out together: since $a_n=\frac{n!}{5^n}$, the ratio is $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{5^{n+1}}}{\frac{n!}{5^n}}=\frac{\frac{(n+1)\cdot n!}{5\cdot 5^n}}{\frac{n!}{5^n}}=\frac{n+1}{5}.$$ Therefore, $$L=\lim_{n\to\infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=\lim_{n\to\infty} \frac{n+1}{5}=\infty,$$ and so by the ratio test we know this series diverges.

We'll end this section with a proof of how the root test can be used to show the ratio test. Suppose we have a series $\sum a_n$ and $$L=\lim_{n\to\infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|.$$ First, we'll show that if $L<1$ then the series converges uniformly.

So, suppose $L<1$- i.e. $$\lim_{n\to\infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|<1.$$ This means that eventually, the ratio of consecutive terms of the sequence will become less than $1$: i.e. there exist a constant $r$ with $0<r<1$ and an integer $N$ such that for any integer $n > N$, we have $$\bigg|\frac{a_{n+1}}{a_n}\bigg|<r<1.$$ We can rewrite this as $$|a_{n+1}|<r|a_n|.$$ So, repeating the process, we can write $$|a_{N+k}|\le r|a_{N+k-1}|\le r\cdot (r|a_{N+k-2}|)\le\cdots\le r^k|a_N|$$ for any $k\ge 1$. Consequently, if $N+k=n$, then $k=n-N$ so we can say that $$|a_n|\le r^{n-N}|a_N|.$$ Taking the $n$-th root of both sides, we get $$|a_n|^{\frac{1}{n}}\le r^{\frac{n-N}{n}}|a_N|^{\frac{1}{n}}=r^{1-\frac{N}{n}}|a_N|^{\frac{1}{n}}.$$ Next, remember that $|a_N|$ is a constant, and so if we take the limit of both sides as $n$ approaches infinity, we get $$\lim_{n\to\infty} |a_n|^{\frac{1}{n}}\le \lim_{n\to\infty} r^{1-\frac{N}{n}}|a_N|^{\frac{1}{n}}=r^{\lim_{n\to\infty} \left(1-\frac{N}{n}\right)}|a_N|^{\lim_{n\to\infty} \frac{1}{n}}=r^1|a_N|^0=r,$$ which is less than $1$ by the definition of $r$. Therefore, the limit on the left is less than $1$, and so by the root test we get that $\sum a_n$ converges.

Next, let's consider the case when $L>1$. In this case, we know that the ratio of consecutive terms of the sequence will become greater than $1$: i.e. there exist a constant $c$ with $c>1$ and an integer $N$ such that for any integer $n > N$, we have $$\bigg|\frac{a_{n+1}}{a_n}\bigg|>c>1.$$ Hence, by the same reasoning as before, we can write $$|a_n|^{\frac{1}{n}}\ge c^{1-\frac{n}{N}}|a_N|^{\frac{1}{n}}.$$ So again, taking the limit of both sides as $n$ approaches $\infty$, we have $$\lim_{n\to\infty} |a_n|^{\frac{1}{n}}\ge \lim_{n\to\infty} c^{1-\frac{n}{N}}|a_N|^{\frac{1}{n}}=c>1.$$ Therefore, by the root test, $\sum a_n$ diverges.

Phew, that was long! The important thing to take from this proof is that this specifically relied on the root test, showing that in a sense the root test is ``stronger'' than the ratio test. Another shocking fact is that the ratio test (by itself) isn't enough to prove the root test!

In practice, we will choose to use either the Ratio Test or the Root Test, not both! They both are automated comparisons to geometric series (so only use them if something is growing/decaying geometrically fast), and tend to work in the same cases. In particular, if one of the tests gives you $1$, then the other one will as well. So don't waste your time trying the other! (There are series that converge by the Root test when the Ratio test is inconclusive, but that requires

• The limit of the ratio in the ratio test doesn't exist.

• A stronger version of the Root test Such a series is $\sum_{n=0}^\infty 2^{(-1)^n-n}$.)

By the Comparison Test, if $a_i<b_i$ for all positive integers $i$ and $\sum_{i \geq 1}^{} b_i$ converges, then $\sum_{i \geq 1}^{} a_i$ also converges.

Note that the Comparison test does not specify the sequence $b_i$. Our goal now is to wisely pick $b_i$ and develop a new test for absolute convergence.

A geometric sequence is one of the simplest type of sequences for which convergence is easily established. For a geometric sequence with first term $a$ and common ratio $r$, the sum $\sum_{k\geq 0} ar^k$ converges if and only if $|r|<1$.

Now, letting $b_i$ be a geometric sequence, we must find a good choice for $a_i$.

Consider a sequence for which $|a_{k+1}|<r|a_k|$ for all $k \geq N$. Then, we have that $$\begin{equation} |a_{N+k}| \leq r |a_{N+k-1}| \leq \ldots \leq r^{k-1}|a_{N+1}|\leq r^k |a_N|. \end{equation}$$

If we let $b_k=|a_N|r^k$, then the series associated with $b_k$ converges if $|r|<1$, because it is geometric. By the comparison test, so does the sum $|a_N|+|a_{N+1}|+\ldots$. Adding a finite amount of terms to a convergent series still results in a convergent series, so we have that $\sum_{j \geq 1} |a_j|$ converges if $|r|<1$.

The idea is now to consider the limit $$\begin{equation} L= \lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|}. \end{equation}$$

If this value is less than $1$, then we can pick an $r$, such that $L<|r|<1$, for which $|a_{k+1}|<r|a_k|$ for all $k \geq N$. In other words, if $L<1$, then the series associated with the sequence $a_k$ converges. This is called the ** Ratio Test**.

We can also prove that the series diverges if $L>1$. This is because if $|a_{n+1}|>|a_n|$ for all $n \geq N$, then $\lim_{n \rightarrow \infty} |a_n|>0$, because $|x| \geq 0$ for all $x$. By the Divergence Test, the series diverges.

The series may diverge or converge. Does this mean that we made an error? Of course not! This just means that, if $L=1$, the Ratio Test is not strong enough to conclude absolute convergence.

Here's an example. Show that $\sum_{n=1}^{\infty} \frac{1}{n!}$ converges.

Solution: Use the Ratio Test! We have that $$\begin{equation} L= \lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|}=\lim_{n \rightarrow \infty} \frac{n!}{(n+1)!}=\lim_{n \rightarrow \infty}\frac{1}{n+1}=0. \end{equation}$$

Since $L<1$, the series converges. In fact, it converges to $e$, which can be proven using the Taylor series for $e^x$.

Overview of the ratio test

Sometimes the root test is hard to use, if $\lim_{n\to\infty} |a_n|^{\frac{1}{n}}$ is hard to evaluate. However, the root test implies a weaker but simpler test called the ratio test. (In other words, the root test works whenever the ratio test works, but the ratio test might be easier to use.) Namely, if the ratio sequence $|a_{n+1}/a_n|$ converges to a number $r<1$, then the series $\sum a_n$ converges absolutely. If the ratio sequence converges to a number $r>1$, then the series $\sum a_n$ diverges. Otherwise, if the limit $\lim_{n\to\infty} |a_{n+1}/a_n|$ either equals $1$ or does not exist, then the test is inconclusive.