17Calculus Infinite Series - Where To Start - Choosing A Test
How do you know which test to use to determine convergence or divergence of an infinite series? Knowing which test to use is a combination of practice and guessing. However, keep in mind that there may be several tests that will work but there is often a best test that is more efficient and quicker than using other tests.
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As you are first learning these tests, here is a hint on how to determine convergence or divergence of a infinite series. Use the tests in this order until one of them tells you that the series converges or diverges:
Group 1. Divergence (nth-Term) Test
Remember that the divergence test can test only for divergence. If the limit goes to zero, you don't know if the series converges or diverges. Said another way, the only two conclusions you can reach from the divergence test are, divergence or inconclusive. If you don't determine divergence, you have to use another test.
Group 2. See if the series matches one of these special forms.
- p-Series
- Geometric Series
- Alternating Series
- Telescoping Series
Group 3. Core Series Tests
- Ratio Test
- Limit Comparison Test or Direct Comparison Test (whichever you find easiest to use)
- Integral Test
- Root Test
The series table is organized in this format.
Once you get some experience, you will find that you can look at a series and determine by what terms are involved, where to start. But this comes only after working many practice problems and deliberate work on your part to discerning patterns. Here are just a few guidelines that may help ( in no special order ). Add more to this list as you learn them.
1. If the series contains a factorial, try the Ratio Test or the Limit Comparison Test.
2. If the series contains a term with n that also has a power of n, try the Root Test.
3. Start with the Alternating Series if the series is alternating.
4. If the series has positive and negative terms, try determining the convergence or divergence of the absolute value of the series and then apply the absolute convergence theorem.
5. If the series looks integrable, try the Integral Test.
6. If you have a polynomial in the denominator that can be factored and no complicated terms in the numerator, try expanding into partial fractions and see if you have a Telescoping Series.
Of course, these guidelines can be used only if you have the option of choosing any test. Make sure you read the problem statement carefully and, if there is any confusion, ask your instructor what they expect.
Here is a video that builds on this discussion. Basically, he has a list of series and he explains his thought process in choosing a test. He doesn't actually work any problems in this video but he does give a great overview on how to look at each series. You will need some actual experience working problems and have some knowledge of all the tests to be able to understand this video.
PatrickJMT - Infinite Series [12min-46secs]
video by PatrickJMT |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
|---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
|---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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