We learned in the last section that, assuming certain conditions are met, the convergence of a series can be tested by comparing that series to a corresponding improper integral. The resulting convergence test was aptly named the Integral Test.
There is one obvious potential drawback to using the integral test, framed in the form of the following question.
What do we do if we can't compute the integral?
One answer is, instead of comparing our series to an improper integral, we could try to compare the series to a series whose convergence behavior is known. In the remainder of these notes we investigate this idea and make it more rigorous.
We begin by considering the series
$$
\sum_{n=1}^{\infty}\dfrac{1}{3^{n}+1}.
$$
This series resembles the convergent geometric series
$$
\sum_{n=1}^{\infty}\dfrac{1}{3^{n}} = \sum_{n=1}^{\infty}\left(\dfrac{1}{3}\right)^{n} = \dfrac{1}{2}
$$
however, it differs by having a $3^{n}+1$ in the denominator. Nonetheless, we suspect that it must converge similarly to the geometric series. In fact, the inequality
$$
\dfrac{1}{3^{n}+1} < \dfrac{1}{3^{n}}
$$
shows that the terms in the given series are smaller than the terms in the geometric series, and thus, the values of the partial sums will be smaller than $\tfrac{1}{2}$. Therefore, the partial sums form a bounded, monotonically increasing sequence (sound familiar? JM) which is convergent. Additionally, the sum of the series is going to be less than the sum of the geometric seires, i.e.,
$$
\sum_{n=1}^{\infty}\dfrac{1}{3^{n}+1} < \dfrac{1}{2}
$$
This progression of reasoning not only leads us to the following test for convergence, but also an outline for how we might prove it.
The Comparison Test¶
Suppose that $\sum a_{n}$ and $\sum b_{n}$ are series with positive terms.
- If $\sum b_{n}$ is convergent and $a_{n} \leq b_{n}$ for all $n$, then $\sum a_{n}$ is also convergent.
- If $\sum b_{n}$ is divergent and $a_{n} \geq b_{n}$ for all $n$, then $\sum a_{n}$ is also divergent.
Proof:¶
Let
$$
s_{n} = \sum_{i=1}^{n}a_{i} \quad\quad t_{n}=\sum_{i=1}^{n}b_{i}\quad\quad t=\sum_{n=1}^{\infty}b_{n}
$$
(1): We first assume that $\sum b_{n}$ is convergent and that $a_{n}\leq b_{n}$ for all $n$. Since both series have positive terms, the sequences $\lbrace s_{n} \rbrace$ and $\lbrace t_{n} \rbrace $ of partial sums are both increasing. We see this for $\lbrace s_{n} \rbrace$ by noting that,
$$
s_{n} \leq a_{n+1} + s_{n} = s_{n+1}.
$$
and for $\lbrace t_{n} \rbrace $ by remembering that because $\sum b_{n}$ is convergent, we have that $t_{n}\to t$, and thus $t_{n}\leq t$ for all $n$.
Next, since $a_{i}\leq b_{i}$ for all $i$, this implies that $s_{n} \leq t_{n}$. Therefore, $s_{n}\leq t$ for all $n$. This means that the sequence of partial sums $\lbrace s_{n} \rbrace$ is increasing and bounded and therefore converges by the Monotonic Sequence Theorem. Thus, $\sum a_{n}$ converges.
(2): We assume that $\sum b_{n}$ is divergent and that $a_{n}\geq b_{n}$ for all $n$. We know that since $\lbrace t_{n} \rbrace$ is increasing, then $t_{n}\to \infty$. But $a_{i}\geq b_{i}$ for all $i$ and so $s_{n} \geq t_{n}$ for all $n$. Thus $s_{n}\to\infty$. Therefore $\sum a_{n}$ diverges.
Note:¶
In order to apply either the Comparison Test or the Limit Comparison Test, you need to compare the given series with a series that is known to be either convergent or divergent. Most of the time we will use the following.
- A $p$-series
$$ \sum_{n=1}^{\infty}\dfrac{1}{n^{p}} \text{ is } \begin{cases} \text{convergent} & \text{for } p > 1\\ \text{divergent} & \text{for } p\leq 1 \end{cases} $$ - A geometric series
$$ \sum_{n=1}^{\infty}ar^{n-1} \text{ is } \begin{cases} \text{convergent} & \text{for } \vert r \vert < 1\\ \text{divergent} & \text{for } \vert r \vert \geq 1 \end{cases} $$
1. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{n-1}{n^{3}+1}
$$
Check your answer
The series is convergent.
Video solution
2. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{k=1}^{\infty}\dfrac{k\sin^{2}(k)}{1+k^{3}}
$$
Check your answer
The series is convergent.
Video solution
3. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{k=1}^{\infty}\dfrac{4^{n+1}}{3^{n}-2}
$$
Check your answer
The series is divergent.
Video solution
4. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{k=1}^{\infty}\dfrac{n!}{n^{n}}
$$
Check your answer
The series is convergent.
Video solution
Now, consider the series
$$
\sum_{n=1}^{\infty}\dfrac{1}{3^{n}-1}.
$$
This series again resembles the convergent geometric series
$$
\sum_{n=1}^{\infty}\dfrac{1}{3^{n}}
$$
As far as the Comparison Test is concerned, the inequality
$$
\dfrac{1}{3^{n}-1} > \dfrac{1}{3^{n}}
$$
is useless since $\sum \left(\dfrac{1}{3}\right)^{n}$ is convergent and $\dfrac{1}{3^{n}-1}>\dfrac{1}{3^{n}}$. However, the new series should behave (converge) similarly to the geometric series. In these cases, we use the following test.
The Limit Comparison Test¶
Suppose that $\sum a_{n}$ and $\sum b_{n}$ are series with positive terms. If
$$
\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}} = c
$$
where $c$ is a finite number and $c>0$, then either both series converge or both diverge.
Proof:¶
Let $m$ and $M$ be positive numbers such that $m < c < M$. Because the ratio $\dfrac{a_{n}}{b_{n}}$ is close to $c$ for large $n$, there is an integer $N$ such that
$$
m < \dfrac{a_{n}}{b_{n}} < M \quad \text{where } n>N
$$
and so
$$
mb_{n} < a_{n} < Mb_{n} \quad \text{where } n > N
$$
Now, if $\sum b_{n}$ converges, so does $\sum Mb_{n}$ because one is merely a constant multiple of the other. Thus $\sum a_{n}$ converges by part (1) of the Comparison Test. Also, if $\sum b_{n}$ diverges, so does $\sum mb_{n}$ and by part (2) of the Comparison Test, we have that $\sum a_{n}$ diverges.
5. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{n^{2} + n + 1}{n^{4}+n^{2}}
$$
Check your answer
The series is convergent.
Video solution
6. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{\sqrt{1+n}}{2+n}
$$
Check your answer
The series is divergent.
Video solution
7. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{n+3^{n}}{n+2^{n}}
$$
Check your answer
The series is divergent.
Video solution
Estimating Sums¶
Similarly to the Remainder Estimate for the Integral Test, if we have used the Comparison Test to show that a series $\sum a_{n}$ converges by comparison with $\sum b_{n}$, then we may be able to estimate the sum $\sum a_{n}$ by comparing remainders. We will denote the remainder of the series $\sum a_{n}$ by
$$
R_{n} = s - s_{n} = a_{n+1} + a_{n+2} + \cdots
$$
and the remainder for the comparison series $\sum b_{n}$ by
$$
T_{n} = t - t_{n} = b_{n+1} + b_{n+2} + \cdots
$$
Since $a_{n} \leq b_{n}$ for all $n$, we have
$$
R_{n} \leq T_{n}
$$
Now, if $\sum b_{n}$ is a $p-$series, then you use the methods from section 11.3 to estimate the sum. If $\sum b_{n}$ is a geometric series, then its sum is the sum of a geometric series and we can find the sum exactly as in section 11.2.
8. Example¶
Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.
$$
\sum_{n=1}^{\infty}\dfrac{1}{5+n^{5}}
$$
Video solution
Discussion¶
Questions? You can ask them below.