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AP Calc BC in a week - Part 6

Introduction

Well, evidently I failed to do it in 5 days. Ah well.

Today we'll be continuing with Series, testing for convergence.

Tests for convergence of Infinite series

Calm down, it's not that bad.

nth term test

This test is just the contrapositive of the first theorem we did last time:

If \(\lim_{n\to\infty}a_n\neq 0\), \(\sum a_n\) diverges

Apparently it's good to start with this test.

If the terms don't approach zero, then the series diverges

Beware that if the terms do approach zero, it doesn't mean that the series converges!

Geometric series test

The geometric series \(\sum ar^n\) converges if and only if \(|r|<1\).

If \(|r|<1\), then the sum is \(\frac{a}{1-r}\)

tl;dr

  • If the terms of the series don't approach zero, then the series diverges
  • If and only if a geometric series has \(|r|<1\), the series converges.

Tests for convergence of Nonnegative series

A nonnegative series is a series where all its terms are \(\ge 0\)

Integral test

I'm going to be non-rigorous here.

\(a_n=f(n)\), so if \(f\) is a continuous, positive, always decreasing function, then then \(\sum a_n\) converges if and only if \(\int_1^\infty f(x)dx\) converges

I suppose it's intuitive? If the infinite integral of your function/series converges then your series converges. Do some practice and review.

tl;dr turn your (always positive) series into an integral and if it converges, your series converges

p-series test

Ooh this one's pretty easy. It's a test that only works on p-series, and if \(p\le 1\) then the series diverges, and if \(p>1\) then the series converges.

This just comes straight from the p-series section. Go check that one out. The harmonic series ( \(p=1\) ) diverges, the p-series where \(p=2\) converges, so you can kinda remember it from that.

Remember that a p-series looks like \(\sum_{n=1}^\infty \frac{1}{n^p}\), where \(p\) is a constant.

Comparison test

I think we did this before.

If a series \(\sum a_n\) is always less than \(\sum u_n\), and we know that \(\sum u_n\) converges, then \(\sum a_n\) also converges.

If a series \(\sum a_n\) is always greater than \(\sum u_n\), and we know that \(\sum u_n\) diverges, then \(\sum a_n\) also diverges.

Apparently p-series and geometric series are good series to compare to, since we can easily know if they will converge or diverge depending on their \(p\) or \(r\).

You should also remember that in order to compare, you can discard a finite number of terms or multiply the series by a nonzero constant and not affect the convergence.

tl;dr series less than converging series converges, series greater than diverging series diverges.

Limit comparison test

If \(\lim{n\to\infty}\frac{a_n}{b_n}\) is finite and nonzero, then \(\sum a_n\) and \(\sum b_n\) both converge or both diverge.

Apparently this test is used if the comparison test fails, for instance (from the book) \(\frac{1}{2n+1}\) seems to be related to the harmonic series but is less than the harmonic series so the comparison test won't work.

If the limit is zero or infinity, try another test.

tl;dr If the limit of series A over series B is finite and nonzero, then the series both converge or both diverge.

Ratio test

If limit \(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\) is \(<1\) then it converges, if it is \(>1\) then it diverges. If it doesn't exist or if it equals 1 then it doesn't work.

I suppose the direction of the comparison could be remembered pretty easily since it's the ratio of the next term over the previous term. If it's \(<1\), then that means that the next term is less than the previous term so it converges. If it's \(>1\), then that means that the next term is greater than the previous term so it diverges.

tl;dr Limit of next term over previous term. If \(<1\), converges, if \(>1\), diverges.

nth root test

I totally do not understand this test so let's memorize it:

If limit \(\lim_{n\to\infty}\sqrt[n]{a_n}\) is \(<1\) then it converges, if it is \(>1\) then it diverges. If it doesn't exist or if it equals 1 then it doesn't work.

The conditions are exactly the same as the ratio test.

I guess this is used for series who are like complicated-expression to the power of \(n\).

Alternating Series and Absolute convergence

I'm going to go ahead and post this blog post before finishing this section, but I'll update it tomorrow with the content, so don't worry.

Alternating Series test


2016-03-29 Paul Elder

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