# Math for Non-Geeks/Direct comparison test

In this chapter we will study an important convergence criterion called the direct comparison test. This criterion allows us to deduce the convergence behavior of a series by comparing it to that of another series. That way we can answer questions regarding the convergence of a series by considering a simpler series. Using this criterion we can estimate upper or lower bounds of the series, until we hopefully find a proof for the convergence behavior.
The direct comparison test is also used to proof other criteria such as the quotient test and the root test, which are both useful tools to have when solving problems about the convergence of series.

## Direct comparison test: majorant

There are two versions of the direct comparison test. The first one is a test for convergence and involves finding a majorant of the series:
Theorem (Direct comparison: majorant)
Let $\sum _{k=1}^{\infty }a_{k}$ be a series. If there exists a convergent series $\sum _{k=1}^{\infty }c_{k}$ with $|a_{k}|\leq c_{k}$ for all $k\in \mathbb {N}$ then $\sum _{k=1}^{\infty }a_{k}$ converges absolutely.
Note that from the inequality $|a_{k}|\leq c_{k}$ it automatically follows that $c_{k}\geq 0$, because $c_{k}$ is greater or equal to the non negative number $|a_{k}|$.
Proof (Direct comparison: majorant)
If $\sum _{k=1}^{\infty }c_{k}$ converges, then the sequences of partial sums is bounded above (see Satz über beschränkte Reihen). Because $|a_{k}|\leq c_{k}$ for all $k\in \mathbb {N}$ also the corresponding sequence of partial sums of $\sum _{k=1}^{\infty }|a_{k}|$ has an upper bound:
$\sum _{k=1}^{n}|a_{k}|\leq \sum _{k=1}^{n}c_{k}\leq \sum _{k=1}^{\infty }c_{k}<\infty$
The sequence of partial sums $\sum _{k=1}^{\infty }|a_{k}|$ is monotonically increasing since $|a_{k}|\geq 0$. We have that the series is monotonic and bounded. Therefore $\sum _{k=1}^{\infty }|a_{k}|$ is convergent.
Comprehension question: Would it suffice for the majorant if there exists $N\in \mathbb {N}$ so that $|a_{k}|\leq c_{k}$ for all $k\geq N$?
Yes. If the sequence of partial sums $\sum _{k=1}^{\infty }c_{k}$ is bounded above, then this is true also for $\sum _{k=N}^{\infty }c_{k}$. Because of
$\sum _{k=N}^{n}|a_{k}|\leq \sum _{k=N}^{n}c_{k}\leq \sum _{k=N}^{\infty }c_{k}<\infty$
also the sequence of partial sums $\sum _{k=N}^{\infty }|a_{k}|$ is bounded above. Since the finitely many summands $a_{1},a_{2},\ldots ,a_{N-1}$ don't change the boundedness, the partial sums of $\sum _{k=1}^{\infty }|a_{k}|$ also have an upper bound. This means that $\sum _{k=1}^{\infty }a_{k}$ converges absolutely.
Hint
As we already mentioned in the introduction, we would like to find a preferably simple convergent series to use as our majorant. Often, we can use the series $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ as our majorant. We could also use a more general harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}$ for $\alpha >1$. Another good option is the convergent geometric series $\sum _{k=0}^{\infty }q^{k}$ for $|q|<1$, for example $\sum _{k=0}^{\infty }\left({\tfrac {1}{2}}\right)^{k}$.

## Direct comparison test: minorant

The second comparison test is similar, but we use it to determine the divergence of a series, and it involves finding a minorant of that series.
Theorem (Direct comparison test: minorant)
Let $\sum _{k=1}^{\infty }a_{k}$ be a series with $a_{k}\geq 0$ for all $k\in \mathbb {N}$. If there exists a divergent series $\sum _{k=1}^{\infty }c_{k}$ with $a_{k}\geq c_{k}\geq 0$ for all $k\in \mathbb {N}$, then also the series $\sum _{k=1}^{\infty }a_{k}$ is divergent.
Proof (Direct comparison test: minorant)
Since $c_{k}\geq 0$ the sequence of partial sums of $\sum _{k=1}^{\infty }c_{k}$ is monotonic. According to our premise this series is divergent, so it must be unbounded. Because $c_{k}\leq a_{k}$ for all $k\in \mathbb {N}$ we have that $\sum _{k=1}^{n}c_{k}\leq \sum _{k=1}^{n}a_{k}$ for all $n\in \mathbb {N}$ and therefore also the sequence of partial sums of $\sum _{k=1}^{\infty }a_{k}$ is unbounded. This implies that $\sum _{k=1}^{\infty }a_{k}$ diverges (every unbounded sequence diverges.)
Hint
Analogue to the majorant condition it sufficies that $a_{k}\geq c_{k}\geq 0$ be true for all $k\geq N$, for a fixed $N\in \mathbb {N}$.
Hint
When using the minorant criterion in practice, often a good choice for the minorant is the divergent harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$. But also the series $\sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}$ for $\alpha <1$ (for example $\sum _{k=1}^{\infty }{\tfrac {1}{\sqrt {k}}}$) make for good minorants. Also the geometric series $\sum _{k=1}^{\infty }q^{k}$ with $q\geq 1$ is suited as a minorant.
Warning
When applying the direct comparison test using a minorant it is necessary that $a_{k}\geq 0$! If we only have that $|a_{k}|\geq c_{k}$ for $a_{k}\in \mathbb {R}$ as in the majorant case, then we cannot follow the divergence of $\sum _{k=1}^{\infty }a_{k}$ from the divergence of $\sum _{k=1}^{\infty }c_{k}$. We can only say that $\sum _{k=1}^{\infty }a_{k}$ doesn't converge absolutely. To see this consider for example the series $\sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{k}}$ and $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ with $a_{k}={\tfrac {(-1)^{k}}{k}}$ and $c_{k}={\tfrac {1}{k}}$. We have ${\tfrac {1}{k}}\leq \left|{\tfrac {(-1)^{k}}{k}}\right|={\tfrac {1}{k}}$, and the harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ diverges. But we can show that $\sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{k}}$ converges according to the Leibniz-Kriterium.

## Examples and exercises
### Direct comparison with majorant

Example (Direct comparison with majorant)
Take the series $\sum _{k=1}^{\infty }{\tfrac {k}{k^{3}+k}}$. The corresponding sequence is $a_{k}={\tfrac {k}{k^{3}+k}}$. If we factor out $k$ in the denominator and reduce we get:
${\frac {k}{k^{3}+k}}={\frac {k}{k(k^{2}+1)}}={\frac {1}{k^{2}+1}}$
The series behaves like $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ and should therefore converge. We can prove this formally using our direct comparison test: From $k^{2}+1\geq k^{2}$ we have that
$\underbrace {\frac {k}{k^{3}+k}} _{=|a_{k}|}={\frac {1}{k^{2}+1}}\leq \underbrace {\frac {1}{k^{2}}} _{=c_{k}}$
With $\sum _{k=1}^{\infty }c_{k}=\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ we have found a majorant that converges. From our direct comparison test it follows that $\sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }{\tfrac {k}{k^{3}+k}}$ also converges.
Math for Non-Geeks: Template:Aufgabe

### Direct comparison with minorant

Example (Direct comparison with minorant)
Let us examine the series $\sum _{k=1}^{\infty }{\tfrac {1}{2k-1}}$. It should grow similar to $\sum _{k=1}^{\infty }{\tfrac {1}{2k}}$. Since the harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ diverges, also the series $\sum _{k=1}^{\infty }{\tfrac {1}{2k}}$ should diverges, and thus also our original series. Formal proof:
We have
$2k-1\leq 2k\iff \underbrace {\frac {1}{2k-1}} _{=a_{k}}\geq \underbrace {\frac {1}{2k}} _{=c_{k}}$
Thus $\sum _{k=1}^{\infty }{\tfrac {1}{2k}}$ is a minorant and diverges. From direct comparison it follows that $\sum _{k=1}^{\infty }{\tfrac {1}{2k-1}}$ is also divergent.
Math for Non-Geeks: Template:Aufgabe

## Corollary: Limit comparison test

For series with positive summands there is another test called the limit comparison test, which we can derive form the direct comparison test:
Theorem (Limit comparison test)
Let $\sum _{k=1}^{\infty }a_{k}$ and $\sum _{k=1}^{\infty }b_{k}$ be series with positive summands, and $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}=c$ with $c>0$, then the series $\sum _{k=1}^{\infty }a_{k}$ converges if and only if $\sum _{k=1}^{\infty }b_{k}$ converges.
Proof (Limit comparison test)
Because $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}=c$ there exists $N\in \mathbb {N}$, so that $|{\tfrac {a_{k}}{b_{k}}}-c|<{\tfrac {c}{2}}$ for all $k\geq N$. This follows from the epsilon definition of convergence with $\epsilon ={\tfrac {c}{2}}$. For all $k\geq N$ we have:
${\begin{aligned}&|{\tfrac {a_{k}}{b_{k}}}-c|<{\tfrac {c}{2}}\\[0.5em]\iff &c-{\tfrac {c}{2}}\leq {\tfrac {a_{k}}{b_{k}}}\leq c+{\tfrac {c}{2}}\\[0.5em]\iff &{\tfrac {1}{2}}c\leq {\tfrac {a_{k}}{b_{k}}}\leq {\tfrac {3}{2}}c\\[0.5em]\iff &{\tfrac {1}{2}}c\cdot b_{k}\leq a_{k}\leq {\tfrac {3}{2}}c\cdot b_{k}\end{aligned}}$
If $\sum _{k=1}^{\infty }a_{k}$ converges, then also $\sum _{k=1}^{\infty }{\tfrac {1}{2}}c\cdot b_{k}$ converges by direct comparison because of the inequality ${\tfrac {1}{2}}c\cdot b_{k}\leq a_{k}$ for all $k\geq N$. We also see that $\sum _{k=1}^{\infty }b_{k}=\sum _{k=1}^{\infty }{\tfrac {2}{c}}\cdot \left({\tfrac {1}{2}}c\cdot b_{k}\right)$. But then also $\sum _{k=1}^{\infty }b_{k}$ converges, because multiplying by a constant factor doesn't affect convergence behaviour.
On the other side assume that $\sum _{k=1}^{\infty }b_{k}$ is convergent, then also $\sum _{k=1}^{\infty }{\tfrac {3}{2}}c\cdot b_{k}$ is convergent (again multiplication by a constant factor). Because of the inequality $a_{k}\leq {\tfrac {3}{2}}c\cdot b_{k}$ for all $k\geq N$ the series $\sum _{k=1}^{\infty }a_{k}$ is convergent by direct comparison.
Comprehension question: What can we deduce from $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}=0$ under the same conditions?
From the convergence of $\sum _{k=1}^{\infty }b_{k}$ we can deduce the convergence of $\sum _{k=1}^{\infty }a_{k}$. If $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}=0$, there exists $N\in \mathbb {N}$ with ${\tfrac {a_{k}}{b_{k}}}\leq 1\iff a_{k}\leq b_{k}$ for all $k\geq N$. Thus if $\sum _{k=1}^{\infty }b_{k}$ is convergent it follows from direct comparison that $\sum _{k=1}^{\infty }a_{k}$ is convergent.
The opposite is not true: If $\sum _{k=1}^{\infty }a_{k}$ converges we cannot conclude that $\sum _{k=1}^{\infty }b_{k}$ is also convergent. As a counterexample consider $\sum _{k=1}^{\infty }\underbrace {\tfrac {1}{k^{2}}} _{=a_{k}}$ and $\sum _{k=1}^{\infty }\underbrace {\tfrac {1}{k}} _{=b_{k}}$ . Here we have
$\lim _{k\to \infty }{\frac {a_{k}}{b_{k}}}=\lim _{k\to \infty }{\frac {\frac {1}{k^{2}}}{\frac {1}{k}}}=\lim _{k\to \infty }{\frac {1}{k}}=0$
But $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ is convergent and $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ is divergent .
Example (Prove convergence using limit comparison test)
Consider the series $\sum \limits _{k=1}^{\infty }{\tfrac {2k^{2}}{6k^{4}-3k}}$. We have
$a_{k}={\frac {2k^{2}}{6k^{4}-3k}}={\frac {k^{2}}{k^{4}}}\cdot {\frac {2}{6-{\frac {3}{k^{3}}}}}={\frac {1}{k^{2}}}\cdot {\frac {2}{6-{\frac {3}{k^{3}}}}}$
With $b_{k}={\tfrac {1}{k^{2}}}$ it follows that
$\lim _{k\to \infty }{\frac {a_{k}}{b_{k}}}=\lim _{k\to \infty }{\frac {2}{6-{\frac {3}{k^{3}}}}}={\frac {2}{6}}={\frac {1}{3}}$
Since $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ is convergent, is follows from the limit comparison test that $\sum \limits _{k=1}^{\infty }{\tfrac {2k^{2}}{6k^{4}-3k}}$ is also convergent .
Hint
As this example shows, the limit comparison test is a convenient way to prove the convergence of a series. Often, computing the limit $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}$ is easier than finding a suiting majorant. In many lectures however the limit comparison test is not discussed, so you are probably not allowed to use it either. Even so you can still use the argumentation of the proof to find the necessary majorant:
For $a_{k}={\tfrac {2k^{2}}{6k^{4}-3k}}$ and $b_{k}={\tfrac {1}{k^{2}}}$ we find that $\lim _{k\to \infty }{\tfrac {a_{k}}{b_{k}}}={\tfrac {1}{3}}$. There must exist $n_{0}\in \mathbb {N}$ with ${\tfrac {a_{k}}{b_{k}}}<2\cdot {\tfrac {1}{3}}={\tfrac {2}{3}}$ for all $k\geq n_{0}$. This is equivalent to $a_{k}\leq {\tfrac {2}{3}}b_{k}$ for all $k\geq n_{0}$. This way we have found a suitable majorant. Because we know that $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}$ converges, also $\sum _{k=1}^{\infty }{\tfrac {2}{3}}{\tfrac {1}{k^{2}}}$ must converge. Now we use the direct comparison test to show that $\sum _{k=1}^{\infty }{\tfrac {2k^{2}}{6k^{4}-3k}}$ is convergent.
Math for Non-Geeks: Template:Aufgabe