Show Mobile Notice
Show All Notes Hide All Notes
You appear to be on a device with a "narrow" screen width (*i.e.* you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-7 : Comparison Test/Limit Comparison Test

In the previous section we saw how to relate a series to an improper integral to determine the convergence of a series. While the integral test is a nice test, it does force us to do improper integrals which aren’t always easy and, in some cases, may be impossible to determine the convergence of.

For instance, consider the following series.

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} + n}}} \]

In order to use the Integral Test we would have to integrate

\[\int_{{\,0}}^{{\,\infty }}{{\frac{1}{{{3^x} + x}}\,dx}}\]

and we're not even sure if it’s possible to do this integral. Nicely enough for us there is another test that we can use on this series that will be much easier to use.

First, let’s note that the series terms are positive. As with the Integral Test that will be important in this section. Next let’s note that we must have \(x > 0\) since we are integrating on the interval \(0 \le x < \infty \). Likewise, regardless of the value of \(x\) we will always have \({3^x} > 0\). So, if we drop the \(x\) from the denominator the denominator will get smaller and hence the whole fraction will get larger. So,

\[\frac{1}{{{3^n} + n}} < \frac{1}{{{3^n}}}\]

Now,

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]

is a geometric series and we know that since \(\left| r \right| = \left| {\frac{1}{3}} \right| < 1\) the series will converge and its value will be,

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} = \frac{1}{{1 - \frac{1}{3}}} = \frac{3}{2}\]

Now, if we go back to our original series and write down the partial sums we get,

\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} \]

Since all the terms are positive adding a new term will only make the number larger and so the sequence of partial sums must be an increasing sequence.

\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} < \sum\limits_{i = 0}^{n + 1} {\frac{1}{{{3^i} + i}}} = {s_{n + 1}}\]

Then since,

\[\frac{1}{{{3^n} + n}} < \frac{1}{{{3^n}}}\]

and because the terms in these two sequences are positive we can also say that,

\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} < \sum\limits_{i = 0}^n {\frac{1}{{{3^i}}} < \sum\limits_{i = 0}^\infty {\frac{1}{{{3^n}}} = \frac{3}{2}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,{s_n} < \frac{3}{2}} } \]

Therefore, the sequence of partial sums is also a bounded sequence. Then from the second section on sequences we know that a monotonic and bounded sequence is also convergent.

So, the sequence of partial sums of our series is a convergent sequence. This means that the series itself,

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} + n}}} \]

is also convergent.

So, what did we do here? We found a series whose terms were always larger than the original series terms and this new series was also convergent. Then since the original series terms were positive (very important) this meant that the original series was also convergent.

To show that a series (with only positive terms) was divergent we could go through a similar argument and find a new divergent series whose terms are always smaller than the original series. In this case the original series would have to take a value larger than the new series. However, since the new series is divergent its value will be infinite. This means that the original series must also be infinite and hence divergent.

We can summarize all this in the following test.

#### Comparison Test

Suppose that we have two series \(\displaystyle \sum {{a_n}} \) and \(\displaystyle \sum {{b_n}} \) with \({a_n},{b_n} \ge 0\) for all \(n\) and \({a_n} \le {b_n}\) for all \(n\). Then,

- If \(\displaystyle \sum {{b_n}} \) is convergent then so is \(\sum {{a_n}} \).

- If \(\displaystyle \sum {{a_n}} \) is divergent then so is \(\sum {{b_n}} \).

In other words, we have two series of positive terms and the terms of one of the series is always larger than the terms of the other series. Then if the larger series is convergent the smaller series must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent. Note as well that in order to apply this test we need both series to start at the same place.

A formal proof of this test is at the end of this section.

Do not misuse this test. Just because the smaller of the two series converges does not say anything about the larger series. The larger series may still diverge. Likewise, just because we know that the larger of two series diverges we can’t say that the smaller series will also diverge! Be very careful in using this test

Recall that we had a similar test for improper integrals back when we were looking at integration techniques. So, if you could use the comparison test for improper integrals you can use the comparison test for series as they are pretty much the same idea.

Note as well that the requirement that \({a_n},{b_n} \ge 0\) and \({a_n} \le {b_n}\) really only need to be true eventually. In other words, if a couple of the first terms are negative or \({a_n}\require{cancel} \cancel{ \le }\,{b_n}\) for a couple of the first few terms we’re okay. As long as we eventually reach a point where \({a_n},{b_n} \ge 0\) and \({a_n} \le {b_n}\) for all sufficiently large \(n\) the test will work.

To see why this is true let’s suppose that the series start at \(n = k\) and that the conditions of the test are only true for for \(n \ge N + 1\) and for \(k \le n \le N\) at least one of the conditions is not true. If we then look at \(\sum {{a_n}} \) (the same thing could be done for \(\sum {{b_n}} \)) we get,

\[\sum\limits_{\,n = k}^\infty {{a_n}} = \sum\limits_{\,n = k}^N {{a_n}} + \sum\limits_{\,n = N + 1}^\infty {{a_n}} \]

The first series is nothing more than a finite sum (no matter how large \(N\) is) of finite terms and so will be finite. So, the original series will be convergent/divergent only if the second infinite series on the right is convergent/divergent and the test can be done on the second series as it satisfies the conditions of the test.

Let’s take a look at some examples.

Example 1 Determine if the following series is convergent or divergent.
\[\sum\limits_{n = 1}^\infty {\frac{n}{{{n^2} - {{\cos }^2}\left( n \right)}}} \]

Show Solution
Since the cosine term in the denominator doesn’t get too large we can assume that the series terms will behave like,

\[\frac{n}{{{n^2}}} = \frac{1}{n}\]

which, as a series, will diverge. So, from this we can guess that the series will probably diverge and so we’ll need to find a smaller series that will also diverge.

Recall that from the comparison test with improper integrals that we determined that we can make a fraction smaller by either making the numerator smaller or the denominator larger. In this case the two terms in the denominator are both positive. So, if we drop the cosine term we will in fact be making the denominator larger since we will no longer be subtracting off a positive quantity. Therefore,

\[\frac{n}{{{n^2} - {{\cos }^2}\left( n \right)}} > \frac{n}{{{n^2}}} = \frac{1}{n}\]

Then, since

\[\sum\limits_{n = 1}^\infty {\frac{1}{n}} \]

diverges (it’s harmonic or the \(p\)-series test) by the Comparison Test our original series must also diverge.

Example 2 Determine if the following series is convergent or divergent.
\[\sum\limits_{n = 1}^\infty {\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}}} \]

Show Solution
This example looks somewhat similar to the first one but we are going to have to be careful with it as there are some significant differences.

First, as with the first example the cosine term in the denominator will not get very large and so it won’t affect the behavior of the terms in any meaningful way. Therefore, the temptation at this point is to focus in on the n in the denominator and think that because it is just an n the series will diverge.

That would be correct if we didn’t have much going on in the numerator. In this example, however, we also have an exponential in the numerator that is going to zero very fast. In fact, it is going to zero so fast that it will, in all likelihood, force the series to converge.

So, let’s guess that this series will converge and we’ll need to find a larger series that will also converge.

First, because we are adding two positive numbers in the denominator we can drop the cosine term from the denominator. This will, in turn, make the denominator smaller and so the term will get larger or,

\[\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}} \le \frac{{{{\bf{e}}^{ - n}}}}{n}\]

Next, we know that \(n \ge 1\) and so if we replace the n in the denominator with its smallest possible value (i.e. 1) the term will again get larger. Doing this gives,

\[\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}} \le \frac{{{{\bf{e}}^{ - n}}}}{n} \le \frac{{{{\bf{e}}^{ - n}}}}{1} = {{\bf{e}}^{ - n}}\]

We can’t do much more, in a way that is useful anyway, to make this larger so let’s see if we can determine if,

\[\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \]

converges or diverges.

We can notice that \(f\left( x \right) = {{\bf{e}}^{ - x}}\) is always positive and it is also decreasing (you can verify that correct?) and so we can use the Integral Test on this series. Doing this gives,

\[\int_{1}^{\infty }{{{{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \int_{1}^{t}{{{{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - {{\bf{e}}^{ - x}}} \right)} \right|_1^t = \mathop {\lim }\limits_{t \to \infty } \left( { - {{\bf{e}}^{ - t}} + {{\bf{e}}^{ - 1}}} \right) = {{\bf{e}}^{ - 1}}\]

Okay, we now know that the integral is convergent and so the series \(\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \) must also be convergent.

Therefore, because \(\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \) is larger than the original series we know that the original series must also converge.

With each of the previous examples we saw that we can’t always just focus in on the denominator when making a guess about the convergence of a series. Sometimes there is something going on in the numerator that will change the convergence of a series from what the denominator tells us should be happening.

We also saw in the previous example that, unlike most of the examples of the comparison test that we’ve done (or will do) both in this section and in the Comparison Test for Improper Integrals, that it won’t always be the denominator that is driving the convergence or divergence. Sometimes it is the numerator that will determine if something will converge or diverge so do not get too locked into only looking at the denominator.

One of the more common mistakes is to just focus in on the denominator and make a guess based just on that. If we’d done that with both of the previous examples we would have guessed wrong so be careful.

Let’s work another example of the comparison test before we move on to a different topic.

Example 3 Determine if the following series converges or diverges.

\[\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4} + 5}}} \]

Show Solution
In this case the “+2” and the “+5” don’t really add anything to the series and so the series terms should behave pretty much like

\[\frac{{{n^2}}}{{{n^4}}} = \frac{1}{{{n^2}}}\]

which will converge as a series. Therefore, we can guess that the original series will converge and we will need to find a larger series which also converges.

This means that we’ll either have to make the numerator larger or the denominator smaller. We can make the denominator smaller by dropping the “+5”. Doing this gives,

\[\frac{{{n^2} + 2}}{{{n^4} + 5}} < \frac{{{n^2} + 2}}{{{n^4}}}\]

At this point, notice that we can’t drop the “+2” from the numerator since this would make the term smaller and that’s not what we want. However, this is actually the furthest that we need to go. Let’s take a look at the following series.

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4}}}} & = \sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{n^4}}}} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^4}}}} \\ & = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^4}}}} \end{align*}\]

As shown, we can write the series as a sum of two series and both of these series are convergent by the \(p\)-series test. Therefore, since each of these series are convergent we know that the sum,

\[\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4}}}} \]

is also a convergent series. Recall that the sum of two convergent series will also be convergent.

Now, since the terms of this series are larger than the terms of the original series we know that the original series must also be convergent by the Comparison Test.

The comparison test is a nice test that allows us to do problems that either we couldn’t have done with the integral test or at the best would have been very difficult to do with the integral test. That doesn’t mean that it doesn’t have problems of its own.

Consider the following series.

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} - n}}} \]

This is not much different from the first series that we looked at. The original series converged because the \(3^{n}\) gets very large very fast and will be significantly larger than the \(n\). Therefore, the \(n\) doesn’t really affect the convergence of the series in that case. The fact that we are now subtracting the \(n\) off instead of adding the \(n\) on really shouldn’t change the convergence. We can say this because the \(3^{n}\) gets very large very fast and the fact that we’re subtracting \(n\) off won’t really change the size of this term for all sufficiently large values of \(n\).

So, we would expect this series to converge. However, the comparison test won’t work with this series. To use the comparison test on this series we would need to find a larger series that we could easily determine the convergence of. In this case we can’t do what we did with the original series. If we drop the \(n\) we will make the denominator larger (since the \(n\) was subtracted off) and so the fraction will get smaller and just like when we looked at the comparison test for improper integrals knowing that the smaller of two series converges does not mean that the larger of the two will also converge.

So, we will need something else to do help us determine the convergence of this series. The following variant of the comparison test will allow us to determine the convergence of this series.

#### Limit Comparison Test

Suppose that we have two series \(\sum {{a_n}} \) and \(\sum {{b_n}} \) with \({a_n} \ge 0,{b_n} > 0\) for all \(n\). Define,

\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\]

If \(c\) is positive (*i.e.* \(c > 0\)) and is finite (*i.e.* \(c < \infty \)) then either both series converge or both series diverge.

The proof of this test is at the end of this section.

Note that it doesn’t really matter which series term is in the numerator for this test, we could just have easily defined \(c\) as,

\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{b_n}}}{{{a_n}}}\]

and we would get the same results. To see why this is, consider the following two definitions.

\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\hspace{0.25in}\hspace{0.25in}\overline{c} = \mathop {\lim }\limits_{n \to \infty } \frac{{{b_n}}}{{{a_n}}}\]

Start with the first definition and rewrite it as follows, then take the limit.

\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\,\,\frac{{{b_n}}}{{{a_n}}}\,\,}} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{{{b_n}}}{{{a_n}}}}} = \frac{1}{{\overline{c}}}\]

In other words, if \(c\) is positive and finite then so is \(\overline{c}\) and if \(\overline{c}\) is positive and finite then so is \(c\). Likewise if \(\overline{c} = 0\) then \(c = \infty \) and if \(\overline{c} = \infty \) then \(c = 0\). Both definitions will give the same results from the test so don’t worry about which series terms should be in the numerator and which should be in the denominator. Choose this to make the limit easy to compute.

Also, this really is a comparison test in some ways. If \(c\) is positive and finite this is saying that both of the series terms will behave in generally the same fashion and so we can expect the series themselves to also behave in a similar fashion. If \(c = 0\) or \(c = \infty \) we can’t say this and so the test fails to give any information.

The limit in this test will often be written as,

\[c = \mathop {\lim }\limits_{n \to \infty } {a_n} \cdot \,\,\frac{1}{{{b_n}}}\]

since often both terms will be fractions and this will make the limit easier to deal with.

Let’s see how this test works.

Example 4 Determine if the following series converges or diverges.

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} - n}}} \]

Show Solution
To use the limit comparison test we need to find a second series that we can determine the convergence of easily and has what we assume is the same convergence as the given series. On top of that we will need to choose the new series in such a way as to give us an easy limit to compute for \(c\).

We’ve already guessed that this series converges and since it’s vaguely geometric let’s use

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]

as the second series. We know that this series converges and there is a chance that since both series have the *3*^{n} in it the limit won’t be too bad.

Here’s the limit.

\[\begin{align*}c & = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{3^n}}}\frac{{{3^n} - n}}{1}\\ & = \mathop {\lim }\limits_{n \to \infty } 1 - \frac{n}{{{3^n}}}\end{align*}\]

Now, we’ll need to use L’Hospital’s Rule on the second term in order to actually evaluate this limit.

\[\begin{align*}c & = 1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{3^n}\ln \left( 3 \right)}}\\ & = 1\end{align*}\]

So, \(c\) is positive and finite so by the Comparison Test both series must converge since

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]

converges.

Example 5 Determine if the following series converges or diverges.

\[\sum\limits_{n = 2}^\infty {\frac{{4{n^2} + n}}{{\sqrt[3]{{{n^7} + {n^3}}}}}} \]

Show Solution
Fractions involving only polynomials or polynomials under radicals will behave in the same way as the largest power of \(n\) will behave in the limit. So, the terms in this series should behave as,

\[\frac{{{n^2}}}{{\sqrt[3]{{{n^7}}}}} = \frac{{{n^2}}}{{{n^{\frac{7}{3}}}}} = \frac{1}{{{n^{\frac{1}{3}}}}}\]

and as a series this will diverge by the \(p\)-series test. In fact, this would make a nice choice for our second series in the limit comparison test so let’s use it.

\[\begin{align*}\mathop {\lim }\limits_{n \to \infty } \frac{{4{n^2} + n}}{{\sqrt[3]{{{n^7} + {n^3}}}}}\frac{{{n^{\frac{1}{3}}}}}{1} & = \mathop {\lim }\limits_{n \to \infty } \frac{{4{n^{\frac{7}{3}}} + {n^{\frac{4}{3}}}}}{{\sqrt[3]{{{n^7}\left( {1 + \frac{1}{{{n^4}}}} \right)}}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^{\frac{7}{3}}}\left( {4 + \frac{1}{n}} \right)}}{{{n^{\frac{7}{3}}}\sqrt[3]{{1 + \frac{1}{{{n^4}}}}}}}\\ & = \frac{4}{{\sqrt[3]{1}}} = 4 = c\end{align*}\]

So, \(c\) is positive and finite and so both limits will diverge since

\[\sum\limits_{n = 2}^\infty {\frac{1}{{{n^{\frac{1}{3}}}}}} \]

diverges.

Finally, to see why we need \(c\) to be positive and finite (*i.e.* \(c \ne 0\) and \(c \ne \infty \)) consider the following two series.

\[\sum\limits_{n = 1}^\infty {\frac{1}{n}} \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \]

The first diverges and the second converges.

Now compute each of the following limits.

\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\centerdot \frac{{{n}^{2}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,n=\infty \hspace{0.5in} \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{2}}}\centerdot \frac{n}{1}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0\]

In the first case the limit from the limit comparison test yields \(c = \infty \) and in the second case the limit yields \(c = 0\). Clearly, both series do not have the same convergence.

Note however, that just because we get \(c = 0\) or \(c = \infty \) doesn’t mean that the series will have the opposite convergence. To see this consider the series,

\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \]

Both of these series converge and here are the two possible limits that the limit comparison test uses.

\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{3}}}\centerdot \frac{{{n}^{2}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0 \hspace{0.5in} \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{2}}}\centerdot \frac{{{n}^{3}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,n=\infty \]

So, even though both series had the same convergence we got both \(c = 0\) and \(c = \infty \).

The point of all of this is to remind us that if we get \(c = 0\) or \(c = \infty \) from the limit comparison test we will know that we have chosen the second series incorrectly and we’ll need to find a different choice in order to get any information about the convergence of the series.

We’ll close out this section with proofs of the two tests.

#### Proof of Comparison Test

The test statement did not specify where each series should start. We only need to require that they start at the same place so to help with the proof we’ll assume that the series start at \(n = 1\). If the series don’t start at \(n = 1\) the proof can be redone in exactly the same manner or you could use an index shift to start the series at \(n = 1\) and then this proof will apply.

We’ll start off with the partial sums of each series.

\[{s_n} = \sum\limits_{i = 1}^n {{a_i}} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{t_n} = \sum\limits_{i = 1}^n {{b_i}} \]

Let’s notice a couple of nice facts about these two partial sums. First, because \({a_n},{b_n} \ge 0\) we know that,

\[\begin{align*}{s_n} & \le {s_n} + {a_{n + 1}} = \sum\limits_{i = 1}^n {{a_i}} + {a_{n + 1}} = \sum\limits_{i = 1}^{n + 1} {{a_i}} = {s_{n + 1}}\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}{s_n} \le {s_{n + 1}}\\ {t_n} & \le {t_n} + {b_{n + 1}} = \sum\limits_{i = 1}^n {{b_i}} + {b_{n + 1}} = \sum\limits_{i = 1}^{n + 1} {{b_i}} = {t_{n + 1}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}{t_n} \le {t_{n + 1}}\end{align*}\]

So, both partial sums form increasing sequences.

Also, because \({a_n} \le {b_n}\) for all \(n\) we know that we must have \({s_n} \le {t_n}\) for all \(n\).

With these preliminary facts out of the way we can proceed with the proof of the test itself.

Let’s start out by assuming that \(\sum\limits_{n = 1}^\infty {{b_n}} \) is a convergent series. Since \({b_n} \ge 0\) we know that,

\[{t_n} = \sum\limits_{i = 1}^n {{b_i}} \le \sum\limits_{i = 1}^\infty {{b_i}} \]

However, we also have established that \({s_n} \le {t_n}\) for all \(n\) and so for all \(n\) we also have,

\[{s_n} \le \sum\limits_{i = 1}^\infty {{b_i}} \]

Finally, since \(\sum\limits_{n = 1}^\infty {{b_n}} \) is a convergent series it must have a finite value and so the partial sums, \({s_n}\) are bounded above. Therefore, from the second section on sequences we know that a monotonic and bounded sequence is also convergent and so \(\left\{ {{s_n}} \right\}_{n = 1}^\infty \) is a convergent sequence and so \(\sum\limits_{n = 1}^\infty {{a_n}} \) is convergent.

Next, let’s assume that \(\sum\limits_{n = 1}^\infty {{a_n}} \) is divergent. Because \({a_n} \ge 0\) we then know that we must have \({s_n} \to \infty \) as \(n \to \infty \). However, we also know that for all \(n\) we have\({s_n} \le {t_n}\) and therefore we also know that \({t_n} \to \infty \) as \(n \to \infty \).

So, \(\left\{ {{t_n}} \right\}_{n = 1}^\infty \) is a divergent sequence and so \(\sum\limits_{n = 1}^\infty {{b_n}} \) is divergent.

#### Proof of Limit Comparison Test

Because \(0 < c < \infty \) we can find two positive and finite numbers, \(m\) and \(M\), such that \(m < c < M\). Now, because \(c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\) we know that for large enough \(n\) the quotient \(\frac{{{a_n}}}{{{b_n}}}\) must be close to \(c\) and so there must be a positive integer \(N\) such that if \(n > N\) we also have,

\[m < \frac{{{a_n}}}{{{b_n}}} < M\]

Multiplying through by \({b_n}\) gives,

\[m{b_n} < {a_n} < M{b_n}\]

provided \(n > N\).

Now, if \(\sum {{b_n}} \) diverges then so does \(\sum {m{b_n}} \) and so since \(m{b_n} < {a_n}\) for all sufficiently large \(n\) by the Comparison Test \(\sum {{a_n}} \) also diverges.

Likewise, if \(\sum {{b_n}} \) converges then so does \(\sum {M{b_n}} \) and since \({a_n} < M{b_n}\) for all sufficiently large \(n\) by the Comparison Test \(\sum {{a_n}} \) also converges.