Comparison Test

Section 6.5 Comparison Test

As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand.

Example 6.49. Convergence by Comparison.

Does \(\ds\sum_{n=2}^\infty {1\over n^2\ln n}\) converge?

Solution

The obvious first approach, based on what we know, is the Integral Test. Unfortunately, we can't compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a \(p\)-series, that is,

\begin{equation*} {1\over n^2\ln n}\lt {1\over n^2}\text{,} \end{equation*}

when \(n\ge3\text{.}\) Since adding up the terms \(\ds 1/n^2\) doesn't get “too big”, the new series “should” also converge. Let's make this more precise.

The series \(\ds\sum_{n=2}^\infty {1\over n^2\ln n}\) converges if and only if \(\ds\sum_{n=3}^\infty {1\over n^2\ln n}\) converges—all we've done is dropped the initial term. We know that \(\ds\sum_{n=3}^\infty {1\over n^2}\) converges. Looking at two typical partial sums:

\begin{equation*} s_n={1\over 3^2\ln 3}+{1\over 4^2\ln 4}+{1\over 5^2\ln 5}+\cdots+ {1\over n^2\ln n} \lt {1\over 3^2}+{1\over 4^2}+ {1\over 5^2}+\cdots+{1\over n^2}=t_n\text{.} \end{equation*}

Since the \(p\)-series converges, say to \(L\text{,}\) and since the terms are positive, \(\ds t_n\lt L\text{.}\) Since the terms of the new series are positive, the \(\ds s_n\) form an increasing sequence and \(\ds s_n\lt t_n\lt L\) for all \(n\text{.}\) Hence the sequence \(\ds \{s_n\}\) is bounded and so converges.

Like the Integral Test, the so-called Comparison Test can be used to show both convergence and divergence. In the case of the Integral Test, a single calculation will confirm whichever is the case. To use the Comparison Test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly.

Example 6.50. Divergence by Comparison.

Does \(\ds\sum_{n=2}^\infty {1\over\sqrt{n^2-3}}\) converge?

Solution

We observe that the \(-3\) should have little effect compared to the \(\ds n^2\) inside the square root. Therefore, we guess that the terms are enough like \(\ds 1/\sqrt{n^2}=1/n\) so that the series should diverge. This analysis leads us to apply the Comparison Test based on the harmonic series. We note that

\begin{equation*} {1\over\sqrt{n^2-3}} > {1\over\sqrt{n^2}} = {1\over n}\text{,} \end{equation*}

so that

\begin{equation*} s_n={1\over\sqrt{2^2-3}}+{1\over\sqrt{3^2-3}}+\cdots+ {1\over\sqrt{n^2-3}} > {1\over 2} + {1\over3}+\cdots+{1\over n}=t_n\text{,} \end{equation*}

where \(\ds t_n\) is 1 less than the corresponding partial sum of the harmonic series because we start at \(n=2\) instead of \(n=1\text{.}\) Since \(\ds\lim_{n\to\infty}t_n=\infty\text{,}\) \(\ds\lim_{n\to\infty}s_n=\infty\) as well. Hence, the given series diverges.

For reference, we summarize the Comparison Test in a theorem.

Theorem 6.51. Comparison Test for Series.

Suppose that \(\ds a_n\) and \(\ds b_n\) are non-negative for all \(n\) and that \(\ds a_n\le b_n\) when \(n\ge N\text{,}\) for some \(N\text{.}\)

If \(\ds\sum_{n=0}^\infty b_n\) converges, then \(\ds\sum_{n=0}^\infty a_n\) also converges.
If \(\ds\sum_{n=0}^\infty a_n\) diverges, then \(\ds\sum_{n=0}^\infty b_n\) also diverges.

Note:

Sometimes, even when the Integral Test applies, comparison to a known series is easier, so it's generally a good idea to think about doing a comparison before doing the Integral Test.
The general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series.

Example 6.52. Applying the Comparison Test.

Does \(\ds\sum_{n=1}^\infty {1\over\sqrt{n^2+3}}\) converge?

Solution

Just as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately,

\begin{equation*} {1\over\sqrt{n^2+3}} \lt {1\over n}\text{,} \end{equation*}

so we can't compare the series directly to the harmonic series. A little thought leads us to

\begin{equation*} {1\over\sqrt{n^2+3}} > {1\over\sqrt{n^2+3n^2}} = {1\over2n}\text{,} \end{equation*}

so if \(\sum 1/(2n)\) diverges then the given series diverges. But since \(\sum 1/(2n)=(1/2)\sum 1/n\text{,}\) Theorem 6.29implies that it does indeed diverge.

Example 6.53. Integral Test or Comparison Test.

Does \(\ds\sum_{n=2}^\infty {|\sin n|\over n^2}\) converge?

Solution

We can't apply the Integral Test here, because the terms of this series are not decreasing. Just as in Example 6.49, however,

\begin{equation*} {|\sin n|\over n^2}\le {1\over n^2}\text{,} \end{equation*}

because \(|\sin n|\le 1\text{.}\) Once again, the partial sums

\begin{equation*} s_n = \sum_{i=2}^{n} \frac{|\sin i|}{i^2} \end{equation*}

are non-decreasing and bounded above by

\begin{equation*} t_n = \sum_{i=2}^n \frac{1}{i^2} \lt L\text{,} \end{equation*}

where \(L\) is the sum of the \(p\)-series \(\ds \sum_{n=2}^{\infty}\frac{1}{n^2}\text{.}\) So the given series converges.

Exercises for Section 6.5.

Exercise 6.5.1.

Determine whether the series converge or diverge.

\(\ds\sum_{n=1}^\infty {1\over 2n^2+3n+5}\)
Answer
converges
Solution

We know that

\begin{equation*} \frac{1}{2n^2+3n+5} \lt \frac{1}{2n^2} \lt \frac{1}{n^2}\text{.} \end{equation*}

Since

\begin{equation*} \sum_{n=0}^{\infty} \frac{1}{n^2} \end{equation*}

is a \(p\)-series with \(p=2\text{,}\) it converges. Therefore, by the Comparison Test, the series

\begin{equation*} \sum_{n=0}^{\infty}\frac{1}{2n^2+3n+5} \end{equation*}

also converges.
\(\ds\sum_{n=2}^\infty {1\over 2n^2+3n-5}\)
Answer
converges
Solution
We notice that for \(n\geq 2\text{,}\) \(3n-5 \geq 0\) and so
\begin{equation*} \frac{1}{2n^2+3n-5} \leq \frac{1}{2n^2} \lt \frac{1}{n^2}. \end{equation*}
Since
\begin{equation*} \sum_{n=2}^{\infty} \frac{1}{n^2} \end{equation*}
is a \(p\)-series with \(p=2\text{,}\) it converges. Therefore, by the Comparison Test, the series
\begin{equation*} \sum_{n=2}^{\infty}\frac{1}{2n^2+3n-5} \end{equation*}
also converges.
\(\ds\sum_{n=1}^\infty {1\over 2n^2-3n-5}\)
Answer
converges
Solution
We notice that for \(n \geq 4\text{,}\)
\begin{equation*} 2n^2-3n-5 \geq \frac{n^2}{2} \end{equation*}
Since
\begin{equation*} \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2} \end{equation*}
is a \(p\) series with \(p=2\text{,}\) it converges. Hence, by the Comparison Test, the series converges.
\(\ds\sum_{n=1}^\infty {3n+4\over 2n^2+3n+5}\)
Answer
diverges
Solution
We notice that, for large enough \(n\text{,}\)
\begin{equation*} \frac{3n+4}{2n^2+3n+5} \geq \frac{1}{n}. \end{equation*}
Since
\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n} \end{equation*}
diverges, so does the given series.
\(\ds\sum_{n=1}^\infty {3n^2+4\over 2n^2+3n+5}\)
Answer
diverges
Solution
We notice that, for large enough \(n\text{,}\)
\begin{equation*} \frac{3n^2+4}{2n^2+3n+5} \geq 1. \end{equation*}
Since
\begin{equation*} \sum_{n=1}^{\infty} 1 \end{equation*}
diverges, so does the given series.
\(\ds\sum_{n=1}^\infty {\ln n\over n}\)
Answer
diverges
Solution
For \(n \gt 1\) , we have
\begin{equation*} \frac{\ln(n)}{n} \geq \frac{1}{n}. \end{equation*}
Since
\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n} \end{equation*}
diverges, so does the given series.
\(\ds\sum_{n=1}^\infty {\ln n\over n^3}\)
Answer
converges
Solution
For all \(n\geq 1\text{,}\) we have that
\begin{equation*} \ln(n) \lt n \implies \frac{\ln(n)}{n^3} \leq \frac{1}{n^2}. \end{equation*}
Since
\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n^2} \end{equation*}
is a \(p\)-series with \(p=2\text{,}\) it converges. Hence, by the Comparison Test, the series converges.
\(\ds\sum_{n=2}^\infty {1\over \ln n}\)
Answer
diverges
Solution
For all \(n\geq 1\text{,}\) we have that
\begin{equation*} \ln(n) \lt n \implies \frac{1}{\ln(n)} \geq \frac{1}{n}. \end{equation*}
Since
\begin{equation*} \sum_{n=2}^{\infty} \frac{1}{n} \end{equation*}
is a \(p\)-series with \(p=1\text{,}\) it diverges. Hence, by the Comparison Test, the series diverges.
\(\ds\sum_{n=1}^\infty {3^n\over 2^n+5^n}\)
Answer
converges
Solution

We use the Comparison Test:

\begin{equation*} \frac{3^n}{2^n+5^n} \lt \frac{3^n}{5^n} = \left(\frac{3}{5}\right)^n\text{.} \end{equation*}

The series

\begin{equation*} \sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^n \end{equation*}

is a convergent geometric series (since \(3/5 \lt 1\)). Thus, the series

\begin{equation*} \sum_{n=0}^{\infty} \frac{3^n}{2^n+5^n} \end{equation*}

converges.
\(\ds\sum_{n=1}^\infty {3^n\over 2^n+3^n}\)
Answer
diverges
Solution
We notice that
\begin{equation*} \frac{3^n}{2^n+3^n} \geq \frac{3^n}{2(3^n)} = \frac{1}{2}. \end{equation*}
Since
\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{2} \end{equation*}
diverges, the given series also diverges.