Comparison Tests

Comparison Test

The first real determiner of convergence is the comparison test. This test is very basic and intuitive.

Comparison for Convergence and Divergence

If two series, $S=\sum _{n=j}^{\infty }{s_{n}}$ and $Z=\sum _{n=j}^{\infty }{z_{n}}$ , and if $0\leq z_{n}\leq s_{n}$ in the interval $[j,\infty )$ , then if

$Z$ is divergent, so is $S$

$S$ is convergent, so is $Z$

First, a few words about this test. Notice that this test applies even if the two series' summands are equal. This is because if summands are the same, this means that the series must also be the same, and so if one of them converges or diverges by the equality property they must both converge or diverge. However, if the starting point $j$ is different from series to series, then they will not converge to the same value, that is to say $Z\neq S$ , but this test will still apply.

The test itself follows from the fact that if we know that $S$ converges to some finite number, and we know that $z_{n}$ is less than (or equal to) $s_{n}$ for all $n$ then it follows that $Z$ should also converge to some finite number greater than zero. i.e., if there is a sum $1+2+3+4$ and a sum $2+3+4+5$ then we know that the first sum will be smaller because it has smaller numbers; the only thing smaller than a finite number is another finite number. The same is true for the divergence portion of this test. If $Z$ diverges and $z_{n}$ is less than or equal to $s_{n}$ for all $n$ then $S$ will diverge for essentially the same reason: the summand is bigger, and the sum of a set of numbers greater than the sum of numbers that is infinite must also be infinite as there is no finite number larger than an infinite number.

One last key note is that all the terms $z_{n}$ and $s_{n}$ must be larger than zero in order for this test to be conclusive. The series $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}$ cannot be tested with the comparison test because it is alternating and half the terms are less than zero.

Example 1

Use the divergent and monotonic harmonic series $\sum _{n=1}^{\infty }{\frac {1}{n}}$ to determine if the following series are divergent or if the test is inconclusive.

$\sum _{n=1}^{\infty }{\frac {1}{n+1}}$
$\sum _{n=2}^{\infty }{\frac {1}{n-1}}$
$\sum _{n=1}^{\infty }{\frac {3}{n}}$
$\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}$
$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$

Solutions

Notice that this sum can be rewritten as $\sum _{n=2}^{\infty }{\frac {1}{n}}$ , making it have the same summand as the harmonic series which is divergent; therefore, this series is divergent.
This series is similar: it can be rewritten as $\sum _{n=1}^{\infty }{\frac {1}{n}}$ which is the harmonic series and so it is divergent.
For each $n$ , this series is larger because $3$ divided by any integer is larger than $1$ divided by any integer. This can also be seen as $3\times \sum _{n=1}^{\infty }{\frac {1}{n}}$ , which is essentially $3\times \infty$ and so this series is divergent.
For every $n$ in this series, the summand of $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}$ is larger than the summand of the harmonic series and so this series is divergent. This can be seen by simply plotting the graph of $f(n)={\frac {1}{\sqrt {n}}}$ . Something interesting to note is that when $n<1$ , the summand of the harmonic series is actually larger.
Via plotting/plugging in values of $n$ , we see that for every $n$ in the series, the summand of the harmonic series is larger and so the test fails and is inconclusive.

Example 2

Use the convergent and monotonic series $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ to determine whether the following series are convergent or if the test is inconclusive.

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$
$\sum _{n=1}^{\infty }{e^{-x}}$
$\sum _{n=1}^{\infty }{{\frac {1}{2^{n}}}\sin ^{2}(x)}$
$\sum _{n=1}^{\infty }{\frac {2}{2^{n}}}$
$\sum _{n=1}^{\infty }{\frac {1}{1.5^{n}}}$
$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{2^{n}}}$

Solutions

$n^{-2}$ decreases at a faster rate than $2^{-n}$ . However, these series do not satisfy the $0\leq z_{n}\leq s_{n}$ requirement, because $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ is larger than $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ when $n<2$ . We can solve this issue by taking removing the first term from the both series to obtain $1+\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ and $2+\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}$ . Now, comparing $\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ with $\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}$ shows that $\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ is indeed convergent. Because this is convergent, adding the original $1$ will not change whether it is convergent or not, it will add $1$ to the value of convergence.
$\sum _{n=1}^{\infty }{e^{-x}}$ is smaller $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ for every $n$ , so this series is convergent.
$\sum _{n=1}^{\infty }{{\frac {1}{2^{n}}}\sin ^{2}(x)}$ is less than or equal to $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ and is greater than $0$ for every $n$ in the domain; this is because $\sin ^{2}(x)$ conforms to </math>\frac{1}{2^n}</math>, and the fact that $\sin(x)$ is squared implies that it will never be less than zero.
$\sum _{n=1}^{\infty }{\frac {2}{2^{n}}}$ is convergent. Notice that this is just $2\times \sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ , which is just $2$ multiplied by some finite number.
$\sum _{n=1}^{\infty }{\frac {1}{1.5^{n}}}$ is greater than $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ for an infinite amount of $n$ so the test in inconclusive.
This series is not greater than or equal to zero for an infinite amount of $n$ 's, so this test is inconclusive.