series Archives - rweber.net https://www.rweber.net/tag/series/ trying to be a mile wide AND a mile deep Fri, 04 Dec 2015 01:06:29 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.4 37896774 Power series special cases https://www.rweber.net/mathematics/calculus/power-series-special-cases/ https://www.rweber.net/mathematics/calculus/power-series-special-cases/#respond Mon, 19 May 2014 12:00:42 +0000 http://www.rweber.net/?p=5525 Using the ease of moving between geometric series and their sums, plus the fact that for power series the derivative of a series is the sum of the derivatives of each term (and likewise for integrals), we can find a wider range of power series without too much trouble. I like making diagrams so here […]

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]]> Using the ease of moving between geometric series and their sums, plus the fact that for power series the derivative of a series is the sum of the derivatives of each term (and likewise for integrals), we can find a wider range of power series without too much trouble. I like making diagrams so here we go.

1. Find a power series representation of \ln(1+x).

This function differentiates to a fraction we can interpret as the sum of a geometric series.

power series 1

To complete Step 4, note that for x=0, \ln(1+x) = 0 and the series also equals 0, so C must be 0. The radius of convergence for the series obtained in Step 2 is 1, with an interval of (-1, 1). When you integrate the interval of convergence can gain or lose endpoints, but that’s a small thing to check after this fairly quick conversion.

2. Find a power series representation of \displaystyle{ \frac{5}{(3-x)^2} } .

Instead of differentiating, this time we integrate to get the sum of a geometric series.

power series 2

In Step 3 note that from the perspective of the derivative operator 3^{n+1} is constant. There’s no constant of integration to find so once you’ve rounded the bases you’re done – well, except for thinking about the radius and interval of convergence.

One more after the jump.

3. Find a power series representation for \displaystyle{\int \tan^{-1}(x) \:dx}.

This time we differentiate twice.

power series 3

Once again you do not have to find C at the end, but in this case it is because C stays undefined: the series is intended to equal an indefinite integral, not a specific function. C does not affect the radius or interval of convergence of the series.

We can use the series above to find a power series for the definite integral \displaystyle{\int_0^1 \tan^{-1}(x) \:dx}. Since plugging 0 into the series obtained in Step 6 gives 0, the answer is the result of plugging in 1: \displaystyle{ \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)} }.

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]]> https://www.rweber.net/mathematics/calculus/power-series-special-cases/feed/ 0 5525 Working with factorial in series https://www.rweber.net/mathematics/calculus/working-factorial-series/ https://www.rweber.net/mathematics/calculus/working-factorial-series/#respond Thu, 08 May 2014 12:00:20 +0000 http://www.rweber.net/?p=5594 To work with factorial in series, you often simply need the ratio test. However, sometimes the ratio test doesn’t give a tractable fraction. In those cases it is good to remember the definition of factorial, in particular the fact that n! = n·(n-1)!, and that tests for convergence typically have conditions that need only hold […]

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]]> To work with factorial in series, you often simply need the ratio test. However, sometimes the ratio test doesn’t give a tractable fraction. In those cases it is good to remember the definition of factorial, in particular the fact that n! = n·(n-1)!, and that tests for convergence typically have conditions that need only hold eventually.

1. For example, the sum from n=1 to infinity of (n!)/(n^n). Once n is at least 2, we can compare to a nice series we know about.

\displaystyle{ \frac{n!}{n^n} \;=\; \frac{n(n-1)\cdots 3\cdot 2\cdot 1}{n\cdot n\cdots n\cdot n\cdot n} \;\leq\; 1\cdot 1\cdots 1\cdot \frac{2}{n}\cdot \frac{1}{n} \;=\; \frac{2}{n^2} }

The sum from n=1 to infinity of 2/(n^2) is easily shown to converge, and, from a finite point on, it sits on top of our original series, which has only positive terms and hence must also converge.

2. We could use the ratio test for the sum \frac{(-3)^n}{n!}, but we can also use the alternating series test. A similar trick to the previous example shows the magnitudes of the terms have limit 0. Once n is at least 3, we have the following comparison on the absolute value of each term.

\displaystyle{ \frac{3^n}{n!} \;=\; \frac{3\cdot 3 \cdots 3 \cdot 3 \cdot 3}{n(n-1)\cdots 3\cdot 2\cdot 1} \;\leq\; \frac{3}{n} \cdot 1 \cdots 1 \cdot \frac{3}{2} \cdot \frac{3}{1} \;=\; \frac{27}{2n} }

That last fraction has a limit of 0 as n goes to infinity.

However, for the alternating series test we need more than a limit of zero; the magnitudes must decrease to 0, for which we use the fact that (n+1)! can be written in terms of n!:

\displaystyle{ \frac{3^{n+1}}{(n+1)!} \;=\; \frac{3}{n+1} \cdot \frac{3^n}{n!} }

As long as n is at least 3, the n+1st term is the nth term times a value less than 1. The series doesn’t decrease right from the start, but from a finite point on it always decreases, and that is enough.

3. How about the sum from n=1 to infinity of \displaystyle{ \frac{(n+2)!}{n!n^2}}? You might be tempted to use the ratio test for this example, but it would waste your time. The terms simplify to [(n+1)(n+2)]/n^2, and have a limit of 1 as n goes to infinity, and therefore the series diverges by the test for divergence.

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]]> https://www.rweber.net/mathematics/calculus/working-factorial-series/feed/ 0 5594 Series convergence https://www.rweber.net/mathematics/calculus/series-convergence/ https://www.rweber.net/mathematics/calculus/series-convergence/#respond Mon, 05 May 2014 12:00:33 +0000 http://www.rweber.net/?p=5584 In a number of tests for series convergence and divergence, you locate or calculate a quantity and draw conclusions based on its value. Here’s a table of which values give what conclusions, for five such tests. Note that the table assumes the series is of the correct form for the test to apply at all […]

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]]> In a number of tests for series convergence and divergence, you locate or calculate a quantity and draw conclusions based on its value. Here’s a table of which values give what conclusions, for five such tests. Note that the table assumes the series is of the correct form for the test to apply at all (although that is only a restriction on p-series and geometric series, in this instance).

Test Value to find Convergent Divergent Inconclusive
test for divergence \lim_{n\to\infty} a_n N/A ≠ 0 = 0
p-series p in  \frac{1}{n^p} > 1 ≤ 1 N/A
geometric series |r| in  ar^n < 1 ≥ 1 N/A
ratio test \lim_{n\to\infty} |a_{n+1}/a_n| < 1 > 1 = 1
root test \lim_{n\to\infty} \sqrt[n]{|a_n|} < 1 > 1 = 1

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]]> https://www.rweber.net/mathematics/calculus/series-convergence/feed/ 0 5584 All roads lead to the ratio test https://www.rweber.net/mathematics/calculus/roads-lead-ratio-test/ https://www.rweber.net/mathematics/calculus/roads-lead-ratio-test/#respond Mon, 17 Mar 2014 12:00:22 +0000 http://www.rweber.net/?p=5537 Happy St. Patrick’s Day! Here, have an awkward flow chart.

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]]> Happy St. Patrick’s Day! Here, have an awkward flow chart.

series flow chart

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]]> https://www.rweber.net/mathematics/calculus/roads-lead-ratio-test/feed/ 0 5537 Musings on Series https://www.rweber.net/mathematics/calculus/musings-series/ https://www.rweber.net/mathematics/calculus/musings-series/#respond Mon, 18 Nov 2013 13:00:19 +0000 http://www.rweber.net/?p=429 1. “Sufficiently large” is an important concept in sequences and series. In essence it means any crazy thing can happen for as long as it wants to happen, so long as there is a finite point after which the sequence or series starts behaving in a controlled/predictable way. A finite number of terms can’t affect […]

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]]> 1. “Sufficiently large” is an important concept in sequences and series. In essence it means any crazy thing can happen for as long as it wants to happen, so long as there is a finite point after which the sequence or series starts behaving in a controlled/predictable way. A finite number of terms can’t affect the limit, and they have a finite sum and so can only affect the series’ value, not whether it converges or not.

2. Advice I’ve given students: If the series does not look like anything but you’re being asked to evaluate it, try partial fractions and see if you get something telescoping.

3. The limit comparison test asks whether the terms of two series are “proportional in the limit.” The ratio and root tests ask, in two ways, whether the terms of one series are “geometric in the limit.” The geometric series with terms cr^n gives ratio r between successive terms (cr^{n+1}/cr^n), leading to the ratio test, and the nth root of its nth term is r times the nth root of c (assuming c is positive, and otherwise taking the negation of the series, which has the same convergence behavior), which limits to r, leading to the root test. This is why the cutoff point for convergence and divergence is 1 – that is what it is for geometric series. The distinction, that at limit 1 we don’t know the behavior, is because this is something only “geometric in the limit.”

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]]> https://www.rweber.net/mathematics/calculus/musings-series/feed/ 0 429 Heavy-handed examples https://www.rweber.net/mathematics/calculus/heavy-handed-examples/ https://www.rweber.net/mathematics/calculus/heavy-handed-examples/#respond Mon, 04 Nov 2013 13:00:50 +0000 http://www.rweber.net/?p=426 I like to present totally cooked-up examples demonstrating the reason for certain hypotheses or limitations in calculus, in the hope that they will help students remember and correctly apply the theorems more easily. Here are a few. Note that the sequence has limit 0 but the function oscillates. This is why you can use convergence […]

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]]> I like to present totally cooked-up examples demonstrating the reason for certain hypotheses or limitations in calculus, in the hope that they will help students remember and correctly apply the theorems more easily. Here are a few.

  • Note that the sequence \sin(2\pi n) has limit 0 but the function \sin(2\pi x) oscillates. This is why you can use convergence of the function to conclude convergence of the sequence of points along its graph, but not vice-versa. Intuitively, the function must “connect the dots” for its behavior to match the sequence’s (but the sequence can never be wilder than the function).
  • Let a_n = n and b_n = -n. These sequences each diverge (to \pm \infty) but their sum is constantly 0. That is why the Limit Laws only allow you to transfer convergence of individual sequences to their sums, products, etc., and not the reverse.
  • Consider the series with terms 1, -1/2, 2/3, -1/3, 1/2, -1/4, 2/5, -1/5, … It is an alternating series whose terms limit to zero. If we sum consecutive pairs of terms, however, we obtain 1/2, 1/3, 1/4, 1/5, …: the harmonic series, which we know diverges. This is why the Alternating Series Test requires the magnitude of the terms decrease to zero.

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]]> https://www.rweber.net/mathematics/calculus/heavy-handed-examples/feed/ 0 426