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Limit Comparison Test

The next test we will be looking at is the limit comparison test; this is not to be confused with the comparison test.

Limit Comparison Test for Convergence and Divergence

If two series $S=\sum _{n=j}^{\infty }{s_{n}}$ and $Z=\sum _{n=j}^{\infty }{z_{n}}$, and if $0\leq z_{n}$ and $0\leq s_{n}$, and $L=\lim _{n\rightarrow \infty }{\frac {s_{n}}{z_{n}}}$ then if $L$ is some positive finite number, either both series diverge or both series converge.

For this test, which term is in the numerator really does not matter. However, neither case will necessarily get you the same answer. The fractions ${\frac {x^{2}+3}{2x}}$ and ${\frac {2x}{x^{2}+3}}$ are reciprocals; taking the limit to infinity in one yields zero and infinity in the other. If the limit is infinity, the test fails. However, taking the limit of the fraction that yields zero will all lead to an inconclusive! Notice that $L$ has to be finite and positive.

To understand why this is the case, let's consider why this test works. If we are given a number $L=\lim _{n\rightarrow \infty }{\frac {s_{n}}{z_{n}}}$ that is assumed to be positive and finite, then there must exist two numbers both smaller and larger than $L$, in essence $k<L<K$. We also know that for some very large $n$ the quotient of the sequences will be very close to $L$, so we can see that $L$ is essentially equal to ${\frac {s_{n}}{z_{n}}}$. This means we can make the substitution $k<{\frac {s_{n}}{z_{n}}}<K$. Multiplying by $z_{n}$ we see that $kz_{n}<s_{n}<Kz_{n}$. Now we have two instances: $kz_{n}<s_{n}$ and $s_{n}<Kz_{n}$. If $Z$ converges then $mZ$ must also converge; the same applies for convergence. So if we know that $mZ$ converges, then $S$ must also converge by the comparison test (recall $s_{n}<Kz_{n}$). The same applies for divergence.

If the limit is not strictly positive and finite, then there does not necessarily exist numbers such that $kz_{n}<s_{n}<Kz_{n}$. (Note multiplication of negative numbers on inequalities.)

Example 1

Use the limit comparison test with $\sum _{n=0}^{\infty }{\frac {1}{3^{n}}}$

1. $\sum _{n=0}^{\infty }{{\left({\frac {2}{3}}\right)}^{n}}$
2. $\sum _{n=0}^{\infty }{\frac {2}{3^{n}}}$
3. $\sum _{n=0}^{\infty }{}$

Definition of a Series

Series

A series is the sum of the terms in a sequence.

For a sequence $d$, the series $D$ would be $D=d_{1}+d_{2}+d_{3}...$. This is true for all series, as it follows from the definition. Only adding a sub-sequence is called a partial sum.

Summation Notation

Purely using the prior definition of a series is possible, but unwieldy. Instead we can again put to use summation notation, which was partially covered in the section on 'integrals'. Some common properties and identities are outlined here.

Identities

$\sum _{k=0}^{n}{c}=nc$ where $c$ is some constant.
$\sum _{k=0}^{n}{k}={\frac {n(n+1)}{2}}$
$\sum _{k=0}^{n}{k^{2}}={\frac {n(n+1)(2n+1)}{6}}$
$\sum _{k=0}^{n}{k^{3}}={\frac {n^{2}(n+1)^{2}}{4}}$

Properties

$\sum _{k}^{n}{s_{k}}+\sum _{n}^{m}{s_{k}}=\sum _{k}^{m}{s_{k}}$ This is the adding of sums.
$j\sum _{k}^{n}{s_{k}}=\sum _{k}^{n}{js_{k}}$ Note that this is essentially the distributive property, so this will work for anything that follows the distributive property, even non-constant terms.

Infinite Series

Infinite Series

An infinite series is the summation of all (or most) terms in an infinite sequence.

An infinite series is just as it sounds: it is an infinitely long sum. A sub-sequence (that is not infinite itself) of an infinite sequence does not form an infinite sequence by definition, so the series created by this sub-sequence would not be an infinite series. However, a sub-sequence composing of a finite fraction of terms of an infinite sequence does form an infinite series, as “half of infinity” is still infinity.

Convergence and Divergence

Convergence of an Infinite Series

An infinite series is said to converge if $\lim _{n\rightarrow \infty }\sum _{k=0}^{\infty }{s_{k}}=L$

Otherwise, it is said to diverge.

Divergence to Infinity of an Infinite Series

An infinite series is said to diverge to infinity if $\lim _{n\rightarrow \infty }\sum _{k=0}^{\infty }{s_{k}}=\infty$

Just as infinite sequences and functions do, infinite series can converge on a specific value. For instance, arithmetic series never converge; instead they diverge to infinity. Some series diverge in different ways. The series

$\sum _{k=0}^{\infty }{(-1)^{k}}$

will diverge by oscillation, as it is constantly fluctuating between 0 and 1. A series can also converge by oscillation, however.

$\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k}}$ does just that.