For most of the series that we'll see, we won't actually find what they converge *to*, but we'll simply get a yes/no answer about whether or not they converge. The tests in this section are how we get those answers (as well as what we've already seen concerning geometric series and others).

The terms must get smaller (approach 0, to be precise) in order for the series to have any hope of converging. \[\text{If } \lim_{n \to \infty} a_n \neq 0 \text{, then } \sum_{n=1}^\infty a_n \text{ is divergent.}\]

Be careful with this test: it doesn't say that if the terms are going to 0, the series is convergent (e.g. the harmonic series). In fact, the divergence test can never conclude that a series converges; it can only ever conclude that a series diverges. For most of the series that we'll see, this test will be inconclusive, but it's an easy one to try, and when it does work, it makes life easy.

Determine whether or not the following series converges. \[\sum_{n=1}^\infty \sin n\]

Since \(\displaystyle\lim_{n \to \infty} \sin n \neq 0\) (this limit doesn't exist)

\[\ans{\sum_{n=1}^\infty \sin n \text{ is divergent, based on the Divergence Test.}}\]Use the Divergence Test to analyze the following series. \[\sum_{n=1}^\infty \dfrac{1}{n^2}\]

Since \(\displaystyle\lim_{n \to \infty} \dfrac{1}{n^2} = 0\)

\[\ans{\text{the Divergence Test is inconclusive.}}\]We need a different test to check this one.

If \(f(x)\) is continuous, positive, and decreasing for all \(x \geq 1\), and \(f(n)=a_n\) for all positive integers \(n\), then \(\displaystyle\sum_{n=1}^\infty a_n\) and \(\displaystyle\int_1^\infty f(x) \ dx\) converge or diverge together.

Use the Integral Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{n+2}{n^2+4n}\]

Start by transitioning to the corresponding function: \[f(x)=\dfrac{x+2}{x^2+4x}.\] This is necessary so that we can integrate. We have to show that this function is continuous, positive, and decreasing:

The two discontinuities in this function occur at \(x=0,-4\), both of which are outside the interval over which we'll be integrating, so we're fine there.

For all \(x \geq 1\), the numerator is positive, and the denominator is positive, so the function as a whole is positive.

To prove that it's decreasing, take the derivative (using the Quotient Rule): \[f'(x)=\dfrac{-x^2-12x-8}{(x^2+4x)^2}\] Since the numerator is negative whenever \(x \geq 1\), and the denominator is always positive, the derivative is negative, meaning that the function is decreasing.

Therefore, this function meets all three criteria, so we can proceed with the Integral Test.

This is an improper integral, and it requires u-substitution. I'll skip some of the details, but you should be able to fill them in.

\[\begin{align} \int_1^\infty \dfrac{x+2}{x^2+4x} \ dx &= \lim_{t \to \infty} \int_1^t \dfrac{x+2}{x^2+4x} \ dx\\ &= \lim_{t \to \infty} \dfrac{1}{2}\ln \bigg|x^2+4x\bigg|_1^t\\ &= \lim_{t \to \infty} \left[\dfrac{1}{2}\ln \big|t^2+4t\big| - \dfrac{1}{2} \ln \big|1+4\big|\right] = \infty\\ \end{align}\]Therefore

\[\ans{\text{by the Integral Test, this series is divergent.}}\]Use the Integral Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{\ln n}{n}\]

This series diverges.

A p-series is a series of the form \(\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^p}\). A p-series is convergent if \(p > 1\), and it is divergent if \(p \leq 1\).

The harmonic series is an example of a divergent p-series, where \(p=1\).

The series \(\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}\) is a convergent p-series, where \(p=2\).

The series \(\displaystyle\sum_{n=1}^\infty \sqrt{n} = \displaystyle\sum_{n=1}^\infty \dfrac{1}{n^{-1/2}}\) is a divergent p-series, where \(p=-1/2\).

If we can compare an unknown series to one that we *do* know about, we may be able to tell whether or not it converges.

Suppose that \(0 < a_n \leq b_n\) for all \(n \geq 1\).

- If \(\displaystyle\sum_{n=1}^\infty b_n\) converges, then \(\displaystyle\sum_{n=1}^\infty a_n\) converges.
- If \(\displaystyle\sum_{n=1}^\infty a_n\) diverges, then \(\displaystyle\sum_{n=1}^\infty b_n\) diverges.

Intuitively, this is similar to the squeeze theorem; if someone in the left lane takes an exit (in the US, at least, where exits are on the right), the person in the right lane is going to be forced to take the exit as well.

This test is most useful when we can compare it to a relatively simple series, like a geometric series or a p-series.

Use the Direct Comparison Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{1}{n^2+4}\]

This series looks like \(\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}\), which is a convergent p-series. Therefore, for the direct comparison test to be conclusive, we need to show that \[\dfrac{1}{n^2+4} \leq \dfrac{1}{n^2}.\]

Since \(n^2+4 > n^2\), it follows that \(\dfrac{1}{n^2+4} < \dfrac{1}{n^2}\). Therefore, since the series in question is bounded above by a convergent series, \[\ans{\sum_{n=1}^\infty \dfrac{1}{n^2+4} \text{ is convergent, based on the Direct Comparison Test.}}\]

Use the Direct Comparison Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{3^n}{2^n-1}\]

This series looks like \(\displaystyle\sum_{n=1}^\infty \dfrac{3^n}{2^n} = \displaystyle\sum_{n=1}^\infty \left(\dfrac{3}{2}\right)^n\), which is a divergent geometric series, since \(r=3/2 > 1\). Therefore, for the direct comparison test to be conclusive, we need to show that \[\dfrac{3^n}{2^n-1} \geq \dfrac{3^n}{2^n}.\]

Since the numerators are equal and \(2^n-1 < 2^n\), it follows that \(\dfrac{3^n}{2^n-1} > \dfrac{3^n}{2^n}\). Therefore, since the series in question is bounded below by a divergent series, \[\ans{\sum_{n=1}^\infty \dfrac{3^n}{2^n-1} \text{ is divergent, based on the Direct Comparison Test.}}\]

What if that last example had been \[\sum_{n=1}^\infty \dfrac{3^n}{2^n+1}\] instead? In that case, the Direct Comparison Test would be inconclusive, since this series would be *below* a divergent series, about which the DCT makes no claim. For something like that, we would need to use the Limit Comparison Test instead.

If \(\displaystyle\sum_{n=1}^\infty a_n\) and \(\displaystyle\sum_{n=1}^\infty b_n\) are series with positive terms and \[\lim_{n \to \infty} \dfrac{a_n}{b_n} = L,\] where \(L\) is **finite** and **nonzero**, then \(\displaystyle\sum_{n=1}^\infty a_n\) and \(\displaystyle\sum_{n=1}^\infty b_n\) converge or diverge together.

Essentially, if this limit is finite and nonzero, the two series track along together, so what one does, the other does. If the limit were infinite, the upper series would be infinitely larger than the lower, and if the limit were zero, the lower series would be infinitely larger than the upper.

Use the Limit Comparison Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{n^3+2n}{n^4-6n^2}\]

This series looks like \(\displaystyle\sum_{n=1}^\infty \dfrac{n^3}{n^4} = \displaystyle\sum_{n=1}^\infty \dfrac{1}{n}\), since the highest-order terms will ultimately dominate the behavior of the series.

\[\lim_{n \to \infty} \dfrac{\dfrac{n^3+2n}{n^4-6n^2}}{\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{n^3+2n}{n^4-6n^2} \cdot \dfrac{n}{1} = \lim_{n=1} \dfrac{n^4+2n^2}{n^4-6n^2} = 1\]Since this limit is finite and nonzero, both series do the same thing. Furthermore, since \(\displaystyle\sum_{n=1}^\infty \dfrac{1}{n}\) is the harmonic series, and thus a divergent p-series, \[\ans{\sum_{n=1}^\infty \dfrac{n^3+2n}{n^4-6n^2} \text{ is divergent, based on the Limit Comparison Test.}}\]

Any time you see a rational expression like this one, the Limit Comparison Test is a good place to start.

The Ratio Test is one of the most important tests, especially for what will come in the following sections, when we deal with Power Series.

For a series \(\displaystyle\sum_{n=1}^\infty a_n\), look at the ratio of subsequent terms \[p = \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|.\]

- If \(p < 1\), the series is convergent.
- If \(p > 1\), the series is divergent.
- If \(p = 1\), the Ratio Test is inconclusive.

Use the Ratio Test to determine whether the following series converges. \[\sum_{n=1}^\infty \dfrac{2^n}{n!}\]

Since \(p < 1\), \[\ans{\sum_{n=1}^\infty \dfrac{2^n}{n!} \text{ is convergent, based on the Ratio Test.}}\]

The Root Test looks similar to the Ratio Test, but it is not nearly so useful. It mainly comes in handy when we have an expression raised to the nth power in a series.

For a series \(\displaystyle\sum_{n=1}^\infty a_n\), look at \[p = \lim_{n \to \infty} \sqrt[n]{|a_n|}.\]

- If \(p < 1\), the series is convergent.
- If \(p > 1\), the series is divergent.
- If \(p = 1\), the Root Test is inconclusive.

Use the Root Test to determine whether the following series converges. \[\sum_{n=2}^\infty \left(\dfrac{1}{\ln n}\right)^n\]

Since \(p < 1\), \[\ans{\sum_{n=2}^\infty \left(\dfrac{1}{\ln n}\right)^n \text{ is convergent, based on the Root Test.}}\]

If a series is an alternating series of the form \[\sum_{n=1}^\infty (-1)^n a_n\] then the series converges if **both** of the following conditions are met:

- \(\displaystyle\lim_{n \to \infty} a_n = 0\) (otherwise the series would fail the Divergence Test)
- \(a_{n+1} \leq a_n\) for all \(n\) greater than some integer \(M\). In other words, these terms need to go to 0 monotonically.

Use the Alternating Series Test to determine whether the following series converges. \[\sum_{n=1}^\infty (-1)^n \cdot \dfrac{n+1}{n^2}\]

Notice that without the alternating part, this series would be divergent (use the Limit Comparison Test with the harmonic series to check that). But since it is alternating, we can show that this one converges, by verifying the two conditions. In fact, the alternating harmonic series converges, even though the basic harmonic series diverges.

- \(\displaystyle\lim_{n \to \infty} a_n = 0\)? It should be clear that \(\dfrac{n+1}{n^2} \to 0\) as \(n \to \infty\).
- \(a_{n+1} \leq a_n\) for all \(n\)? Similarly, since the denominator has \(n^2\) in it, you should be able to tell that as \(n\) increases, the whole expression decreases. To prove this, you can set up the inequality and simplify until you get an obviously true statement.

Since both conditions are met, \[\ans{\sum_{n=1}^\infty (-1)^n \cdot \dfrac{n+1}{n^2} \text{ is convergent, based on the Alternating Series Test.}}\]

So there you have it: there are eight tests in our toolbox, and when you get to a problem and you have to determine whether or not a series is convergent, how do you decide what to do? Of course, you could try all eight tests and see what works, but that would be horribly tedious. So here I give you some guidelines about what to try *first*; of course, more than one test may work on a specific problem, but at least if you have somewhere to start, you can potentially save yourself a few trips down a blind alley.

- Always try the Divergence Test first. Check whether the series in question is a geometric series (or close to one, in which case you can try one of the comparison tests): \[\sum_{n=1}^\infty ar^{n-1}\]
- Check to see if the series is a telescoping series: \[\sum_{n=1}^\infty \dfrac{1}{n} - \dfrac{1}{n+C}\]
- Check whether the series is a p-series (again, if it's close to one, try one of the comparison tests).
- In general, the Limit Comparison Test is more reliable than the Direct Comparison Test. If you see a rational expression, try the LCT.
- If you see a factorial or some piece raised to the nth power, think of the Ratio Test.
- If the whole expression is raised to the nth power, think of the Root Test (or possibly it's a geometric series).
- If it's an alternating series, try the Alternating Series Test.
- If all else fails, try the Integral Test. It's usually more tedious to run than the others, so we save it for a last resort.

- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{2n^2+3n}{n^3+4}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{\sqrt[5]{n}}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{9^n}{13^n}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{n^2 \ 2^{n-1}}{(-5)^n}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{n^2}{n^3+1}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{\sqrt{n+6}}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{n}-\dfrac{1}{n+3}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{2n+3}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{6^n}{n!}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{3^n+4}{2^n}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty (-1)^n \cdot \dfrac{n}{n^2+1}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \left(\dfrac{2n}{3n+7}\right)^n\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{3^n \ n^2}{n!}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{n^4}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{n^2}{10n^2-12}\]
- Determine whether the following series converges or diverges. \[\sum_{n=1}^\infty \dfrac{1}{n(\ln n)^2}\]

Divergent, by the Limit Comparison Test (compare to the harmonic series).

Divergent p-series.

Convergent geometric series (could also use the Root Test).

Convergent, by the Ratio Test.

Divergent, by the Limit Comparison Test (compare to the harmonic series).

Divergent, by the Limit Comparison Test (compared to \(\displaystyle\sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}\)) or the Integral Test.

Convergent Telescoping Series.

Divergent, by the Limit Comparison Test (compare to the harmonic series) or the Integral Test.

Convergent, by the Ratio Test.

Divergent: it is the sum of two geometric series, one of which is divergent.

Convergent, by the Alternating Series Test.

Convergent, by the Root Test.

Convergent, by the Ratio Test.

Convergent p-series.

Divergent, by the Divergence Test.

Divergent, by the Integral Test.