_n=k^

Comparison Test
Examples

When using the comparison test, you have to try and find a suitable series to which you compare the one being tested. The comparison test requires making an educated guess as to whether or not the series will converge. If you think the series will converge, you have to find a bigger series that also converges. If you think the series will diverge, you have to find a smaller series that converges.

We will frequently be making use to two arithmetic properties of sequences:

$$\begin{eqnarray*} \sum_{n=k}^{\infty} (a_n + b_n) & = & \sum_{n=k}^{\infty} a_n ... ... [.1in] \sum_{n=k}^{\infty} c a_n & = & c\sum_{n=k}^{\infty} a_n \end{eqnarray*}$$

Thus, if two series converge, then their sum converges (what if one of the series converged, and the other diverged?), and if you multiply a series by a constant, you do not affect whether or not it converges. Moreover, if $\sum_{n=k}^{\infty} a_n = A$, then $\sum_{n=k}^{\infty} c a_n = cA$, which is exactly what we should expect.

Here is the statement of the comparison test:

Comparison test: Let c_n be a sequence of nonnegative numbers.

• If $\sum c_n$ converges, and $0 \leq a_n \leq c_n$ for all (large enough) n, then $\sum a_n$ converges.
• If $\sum c_n$ diverges, and $a_n \geq c_n$ for all (large enough) n, then $\sum a_n$ diverges.

Now, in order to compare two series using the comparison test, it would be nice to know a few convergence results to compare with.

• p-series: $\sum_{n=k}^{\infty}\frac{1}{n^p}$ converges if and only if p > 1.
  • Note: We will make frequent use of the special case for p=1 (called the Harmonic Series), since it a simple, divergent sequence.
• Geometric Series: $\sum_{n=k}^{\infty}a r^n = \frac{ar^k}{1-r}$ (This is essentially the result you are asked to prove in problem 4, modified slightly to accomodate a lower index of k instead of 0)

Now let's look at a few examples of using the comparison test.

1. Determine whether the series $\sum_{n=k}^{\infty}\frac{1}{n!}$ converges or diverges.

First we need to guess whether the series converges or diverges. Well, since n! grows pretty fast, it seems like the terms $\frac{1}{n!}$ will go to zero pretty quickly, so perhaps the series converges. Let's try comparing it to the series $\sum_{n=k}^{\infty} \frac{1}{n^2}$.

Now, if n! > n² for sufficiently large n, then we will be able to say $\frac{1}{n!} < \frac{1}{n^2}$ (the direction of inequality reverses when you take reciprocals).

So is it true that n! > n² as n gets large? Well, if n>3, we can say

$$n! = n(n-1)(n-2)...3\cdot 2 \cdot 1 \geq n(n-1)(n-2).$$

because a product of numbers that are all greater than or equal to 1 is certainly greater than or equal to a product of any three factors.

So

$$n! \geq n(n-1)(n-2) = n^3 - 3n^2 + 2n > n^2, \textrm{ for sufficiently large } n,$$

because any kth degree polynomial will eventually grow larger than any (k-1)th degree (or lower) polynomial.

We have shown that as n gets larger, eventually n! > n² (if you check, you'll see this holds for $n\geq4$). So we can then deduce that, for sufficiently large n,

$$0 < \frac{1}{n!} < \frac{1}{n^2}$$

.

Therefore, we have satisfied the hypotheses of the comparison test, and since I gave you the information regarding p-series, we know that $\sum_{n=k}^{\infty} \frac{1}{n^2}$ converges. Hence, by the comparison test, so does $\sum_{n=k}^{\infty}\frac{1}{n!}.$

Having shown that $\sum_{n=k}^{\infty}\frac{1}{n!}$ converges, we are now free to use that fact in other comparison test problems.

2. Determine whether the series $\sum_{n=k}^{\infty}\frac{3\sin^2{n}}{n!}$ converges or diverges.

Now that we know $\sum_{n=k}^{\infty}\frac{1}{n!}$ converges, picking a candidate to compare this series to should be simple...

We make use of the fact that $0 \leq \vert\sin{n}\vert \leq 1,$ and therefore $0 \leq \sin^2{n} \leq 1,$ (absolute value bars are not necessary, since we're squaring the sine function)

$$0 < \frac{3\sin^2{n}}{n!} \leq \frac{3}{n!}, (\textrm{ for all }n)$$

And that's about all we need to show. Since constant multiples of convergent series are convergent (i.e. $\sum_{n=k}^{\infty}\frac{3}{n!} = 3 \sum_{n=k}^{\infty}\frac{1}{n!}$), we've satisfied the comparison test hypotheses. Hence $\sum_{n=k}^{\infty}\frac{3\sin^2{n}}{n!}$ converges.

Let's do another one.

3. Determine whether the series $\sum_{n=k}^{\infty}\frac{1}{2\sqrt{n} - 1}$ converges or diverges.

Well, as we look at this series, the most pertinent feature seems to be that $\sqrt{n}$ in the denominator. Ignoring the other numbers for a moment, it seems similar to the series $\sum_{n=k}^{\infty}\frac{1}{\sqrt{n}}$, which is a p-series with $p = \frac{1}{2}$. Since $p \leq 1$ in this case, $\sum_{n=k}^{\infty}\frac{1}{\sqrt{n}}$ diverges. Let's see if we can use this fact to work out the problem at hand.

Let's try and compare this series with $\sum_{n=k}^{\infty}\frac{1}{2\sqrt{n}}$. After all, if $\sum_{n=k}^{\infty}\frac{1}{\sqrt{n}}$ diverges, then so must $\frac{1}{2}\sum_{n=k}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=k}^{\infty}\frac{1}{2\sqrt{n}}$.

We make the casual observation that

$$2\sqrt{n} - 1 < 2\sqrt{n}$$

hence

$$\frac{1}{2\sqrt{n} - 1} > \frac{1}{2\sqrt{n}} \geq 0$$

So we have satisfied the divergence hypothesis of the comparison test. Since $\sum_{n=k}^{\infty}\frac{1}{2\sqrt{n}}$ diverges, so must $\sum_{n=k}^{\infty}\frac{1}{2\sqrt{n} - 1}$.

Finally, it's interesting that this problem gets a little trickier if you change $\sum_{n=k}^{\infty}\frac{1}{2\sqrt{n} - 1}$ to $\sum_{n=k}^{\infty} \frac{1}{2\sqrt{n} + 1}$ (the -1 in the denominator changes to a +1). Now, the first inequality

$$2\sqrt{n} + 1 > 2\sqrt{n}$$

is pointing the wrong way, because when you take reciprocals, you get

$$\frac{1}{2\sqrt{n} + 1} < \frac{1}{2\sqrt{n}}.$$

Showing that a series is less than a divergent series (or greater than a convergent series) doesn't tell you anything at all. There's an easy solution though. Instead of comparing with $\sum_{n=k}^{\infty} \frac{1}{2\sqrt{n}},$ just bump the coefficient of $\sqrt{n}$ in the denominator up to 3; i.e. use the series $\frac{1}{3\sqrt{n}}$. For large enough n, $3\sqrt{n}$ will be greater than $2\sqrt{n} + 1$ ( or $2\sqrt{n} + c$ for that matter, for any constant c).

Hence, adding or subtracting constant terms to polynomials in the denominator doesn't really affect convergence or divergence at all.