Week 4 - Alternating Series

1. Absolutely Convergent Series

1.1. Theorem

  • The series an\sum a_n converges absolutely if the series an\sum \lvert a_n \lvert converges.

1.2. Examples

  • (1)nn2\sum \frac{(-1)^n}{n^2}
    • (1)nn2=1n2\sum \lvert \frac{(-1)^n}{n^2} \rvert = \sum \frac{1}{n^2}
    • And we have proven that 1n2\sum \frac{1}{n^2} converges.
    • So (1)nn2\sum \frac{(-1)^n}{n^2} converges absolutely.

1.3. Procedures to Analyze Series

  • Check limnan\lim_{n \to \infty} a_n,
    • if 0 \ne 0 then diverge.
    • else if an0a_n \ge 0, apply usual convergence tests, like Ratio Test, Root Test and Integral Test.
    • else: Apply tests to an\sum \lvert a_n \rvert. Try to prove the series convergence.

2. Conditional Convergence

2.1. Theorem

  • an\sum a_n is conditional convergence,
    • if an\sum a_n converges,
    • but an\sum \lvert a_n \rvert diverges.

2.2. Examples

  • (1)nn\sum \frac{(-1)^n}{n}
    • (1)nn=1n\sum \lvert \frac{(-1)^n}{n} \rvert = \sum \frac{1}{n} diverges.
    • But it converges, because (1)nn=log2\sum \frac{(-1)^n}{n} = - \log 2

3. Alternating Series

3.1. Definition

  • An alternating series is a series whose terms are alternately positive and negative. Here are some examples:
    • (1)n+1n\sum \frac{(-1)^{n+1} } {n} is alternating series.
      • = 1112+1314+15\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots
    • (1)nsinnn\sum \frac{(-1)^{n} \cdot \sin n } {n} IS NOT.
    • (1)nsin2nn\sum \frac{(-1)^{n} \cdot \sin^2 n } {n} is.

3.2. Alternating Series Test

Theorem

  • If the alternating series n=1(1)n1bn=b1b2+b3b4+b5bn>0\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + b_5 - \cdots b_n > 0 satisfies bn+1bn for all nlimnbn=0\begin{aligned}& b_{n+1} \le b_n \ \text{for all}\ n \\ & \lim_{n \to \infty} b_n = 0\end{aligned},
  • Then, the series converges.

Proven

  • From this figure, we can get:
    • s2ns_{2n} is increasing and bounded above by s1s_1, so limns2n\lim_{n \to \infty} s_{2n} exists.
    • s2n1s_{2n-1} is decreasing and bounded below by s2s_2, so limns2n1\lim_{n \to \infty} s_{2n-1} exists.
  • And we know limn(s2ns2n1)=limn(b2n)\lim_{n \to \infty}(s_{2n} - s_{2n-1}) = \lim_{n \to \infty}(-b_{2n}), and our conditions already says limn(bn)=0\lim_{n \to \infty}(b_{n}) = 0
  • So, limn(s2n)limn(s2n1)=0\lim_{n \to \infty}(s_{2n}) - \lim_{n \to \infty}(s_{2n-1}) = 0, then we can say limn(sn)\lim_{n \to \infty}(s_{n}) also exits and =limn(s2n)=limn(s2n1)=s= \lim_{n \to \infty}(s_{2n}) = \lim_{n \to \infty}(s_{2n-1}) = s.
  • And also, we know, alternating series, is between neighboring partial sums: s2n1ss2ns_{2n-1} \le s \le s_{2n}

Examples

  • Use the fact: alternating series, is between neighboring partial sums: s2n1ss2ns_{2n-1} \le s \le s_{2n} to prove ee is irrational.
    • Definition of Irrational: A real number x is irrational if it cannot be expressed as pq\frac{p}{q} for p, q are integers.
  • Let's declare 1e=(1)nn!\frac{1}{e} = \sum \frac{(-1)^n}{n!} (will learn it in Week 6), so our goal is, to prove (1)nn!\sum \frac{(-1)^n}{n!} irrational.
  • (1)nn!\sum \frac{(-1)^n}{n!} is an alternating series. so sb<1e<sb+1s_b < \frac{1}{e} < s_{b+1}
  • So we can get 0<1esb<1(b+1)!0 < \lvert \frac{1}{e} - s_b \rvert < \frac{1}{(b+1)!} base on sb+1sb=1(b+1)!s_{b+1} - s_b = \frac{1}{(b+1)!}
  • Then 0<b!1esb<b!(b+1)!=1b+10 < b! \cdot \lvert \frac{1}{e} - s_b \rvert < \frac{b!}{(b+1)!} = \frac{1}{b+1}
  • Assume we can rewrite 1e\frac{1}{e} to ab\frac{a}{b}: 0<b!abb!sb<1b+1<10 < \lvert b! \cdot \frac{a}{b} - b! \cdot s_b \rvert < \frac{1}{b+1} < 1
  • We know b!abb! \cdot \frac{a}{b} is an integer. And b!sb=b!n=0b(1)n+1n!=n=0b(1)n+1b!n!b! \cdot s_b = b! \sum_{n=0}^{b} \frac{(-1)^{n+1} }{n!} = \sum_{n=0}^{b} \frac{(-1)^{n+1} \cdot b!}{n!}
    • b is at least as big as n, so n=0b(1)n+1b!n!\sum_{n=0}^{b} \frac{(-1)^{n+1} \cdot b!}{n!} is an integer too.
  • So b!abb!sb\lvert b! \cdot \frac{a}{b} - b! \cdot s_b \rvert should be an integer too. But 0<b!abb!sb<10 < \lvert b! \cdot \frac{a}{b} - b! \cdot s_b \rvert < 1. That's a contradiction.
  • So ee must be irrational.

4. Limit Comparison Test

4.1. Theorem

  • If an0a_n \ge 0, bn0b_n \ge 0 and limnanbn=L>0\lim_{n \to \infty} \frac{a_n}{b_n} = L > 0,
  • Then, bn\sum b_n converges, if and only if an\sum a_n converges.

5. One of the Practice Quiz

  • You may remember that ddxarctan(x)=11+x2.\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}. By computing some terms of the (alternating!) Taylor series for arctangent, approximate arctan(12)\arctan(\frac{1}{2}) to within 133\frac{1}{33}

    • From the formula for geometric series shows that 1+y+y2+y3+=11yif y<1.1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if } \lvert y \rvert \lt 1.
    • And we know x=12<1x = \frac{1}{2} < 1, so we can plug in x2-x^2 for yy, we get that 11+x2=11(x2)=1+(x2)+(x2)2+(x2)3++(x2)n+=1x2+x4x6+x8x10+\begin{aligned} \frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\ &= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\ &= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots \end{aligned}
    • So we have ddxarctan(x)=1x2+x4x6+x8x10+if x<1\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1
    • Then arctan(x)=(ddxarctan(x))dx=(1x2+x4x6+x8x10+)dx=(n=0(1)nx2n)dx=n=0((1)nx2ndx)=n=0((1)nx2ndx)=n=0((1)nx2n+12n+1)+C=C+(xx33+x55x77+x99x1111+).\begin{aligned} \arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\ &= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\ &= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\ &= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1} }{2n+1}\right) + C\\ &= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11} }{11} +\cdots\right). \end{aligned}
    • Evaluating at x=0x = 0 gives 0=arctan(0)=C0 = \arctan(0) = C, so we get arctan(x)=xx33+x55x77+x99x1111+,if x<1.\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11} }{11} + \cdots,\qquad\text{if }|x|\lt 1.
    • x55<133\frac{x^5}{5} < \frac{1}{33}, so the approximate value of arctan(12)=12124=1124\arctan(\frac{1}{2}) = \frac{1}{2} - \frac{1}{24} = \frac{11}{24}
  • Note we are assuming x<1|x| < 1, but arctangent is defined for all real numbers. The series we have here is n=0(1)nx2n+12n+1.\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1} }{2n+1}.

  • Using the Ratio Test, we have that limnan+1an=limnx2n+32n+3x2n+12n+1=limn(2n+1)x2n+3(2n+3)x2n+1=limnx2(2n+1)2n+3=x2limn2n+12n+3=x2.\begin{aligned} \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3} }{2n+3}\quad}{\frac{|x|^{2n+1} }{2n+1} }\\ &= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3} }{(2n+3)|x|^{2n+1} }\\ &= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\ &= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\ &= |x|^2. \end{aligned}
  • By the Ratio Test, the series converges absolutely if x2<1|x|^2 < 1 (that is, if x<1|x| < 1) and diverges if x>1|x| > 1. At x=1x=1 and x=1x=-1, the series is known to converge. So the radius of convergence is 1, and the equality is valid for x[1,1]x \in [-1,1] only.
  • This answer is copied from https://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots.

6. Rearrangement Theorem

  • Let L be a real number, an\sum a_n is a conditional convergent.
  • Then ana_n can be rearranged to form bnb_n and bn=L\sum b_n = L.

6.1. Example

  • (1)n+1n=1112+1314+1516+17\sum \frac{(-1)^{n+1} }{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \cdots
  • Take the even terms, we get 12n=121n\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}, which diverges.
  • Take the odd terms, we get 12n1>12n1\sum \frac{1}{2n-1} > \sum \frac{1}{2n-1}, which diverges too.
  • So whichever we picked, can easily get a number L as large as we want to.

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