Comparison and the Ratio Test

Our most basic tests for convergence are based on the comparison test:

Theorem 4.1 (Comparison Test)   If $0 \leq a_n \leq b_n$ for all n and $\sum b_n$ converges, then so does $\sum a_n$.

PROOF:

Since the terms a_n are non-negative, the sequence of partial sums

$$S_k=\sum_{n=0}^k a_k$$

is monotone nondecreasing. To show that $\sum a_n$ converges we need only give an upper bound for all of the partial sums. Now since $a_n\leq b_n$ for all n,

$$\sum_{n=0}^k a_n \leq \sum_{n=0}^k b_k \leq \sum_{n=0}^\infty b_n$$

Using the convergence of bounded monotone sequences, this tells us that $\sum a_n$ converges, though it does not tell us what it converges to.

Corollary 4.2   If $\sum \vert a_n\vert$ converges then so does $\sum a_n$.

PROOF:

Observe that

$$0\leq a_n+\vert a_n\vert \leq 2\vert a_n\vert.$$

First note that if $\sum \vert a_n\vert$ converges to S then $\sum 2\vert a_n\vert$ will converge to 2S, since each partial sum is doubled. The comparison theorem then tells us that if $\sum 2\vert a_n\vert$ converges then so will $\sum (a_n + \vert a_n\vert)$. Subtracting $\sum \vert a_n\vert$ then tells us that $\sum a_n$ converges.

Definition 4.1   A series $\sum a_n$ is called absolutely convergent if $\sum \vert a_n\vert$ converges.

Since adding a finite sum to the beginning of a convergent series does not change the convergence, though it might change the first M terms, all that matters for determining convergence of a series is what happens in the tail, terms from a_M on for any given M. We'll use this to provide the following comparison with geometric series:

Theorem 4.3 (Ratio Test)   Suppose that a_n>0 for all n and

$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = L.$$

Then if L<1 then $\sum a_n$ converges. If L>1 then the series diverges; if L=1 we don't know what happens.

PROOF:

We will prove the case L<1 by giving a comparison with a suitable geometric series. Since L<1 we can find an r with L<r<1 and for some M if $n\geq M$ then

$$\frac{a_{n+1}}{a_n} < r.$$

We will show that the sequence $${\sum_{n=M}^\infty a_n}$$ converges by comparison with the geometric series $${\sum_{n=0}^\infty a_M r^n}$$. Certainly $a_M \leq a_M r^0$. For $n\geq M$ we have a_n+1< r a_n so using an induction hypothesis we get a_n+1 < r^n+1-Ma_M. This completes the comparison, so $\sum a_n$ converges.

The proof for divergence if L>1 gives a comparison the other way with a divergent geometric series. For L=0 there are examples of both convergent and divergent series (p-series, which make their appearance later in these notes).

Example:

$${\sum \frac{(n!)^2}{(2n)!}}$$ converges. We look at

$$\begin{eqnarray*}\lim_{n\to \infty} \frac{\left(\frac{((n+1)!)^2}{(2n+2)!}\right... ...&=& \lim_{n\to \infty}\frac{(n+1)^2}{(2n+2)(2n+1)} \\ &=& \frac14\end{eqnarray*}$$

Since $L=\frac14 < 1$ this tells us that this series converges.

Exercises: Try out the ratio test on the following series:

1.

$${\sum \frac{4^n}{n!}}$$

2.

$${\sum \frac{n}{2^n}}$$

3.

$${\sum \frac{n^{300}}{1.001^n}}$$

4.

$${\sum \frac{n^2}{n!}}$$

5.

$${\sum \frac{n^4}{4^n}}$$

6.

$${\sum \frac{4^n}{n^3}}$$

7.

$${\sum \frac{n!}{n^n}}$$ Hint: $${\left(\frac{n+1}{n}\right)^n\to e}$$