Week 4 - Alternating Series

Week 4 - Alternating Series #

Absolutely Convergent Series #

Theorem #

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

Examples #

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

Procedures to Analyze Series #

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

Conditional Convergence #

Theorem #

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

Examples #

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

Alternating Series #

Definition #

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

Alternating Series Test #

Theorem #

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

Proven #

  • From this figure, we can get:
    • $s_{2n}$ is increasing and bounded above by $s_1$, so $\displaystyle \lim_{n \to \infty} s_{2n}$ exists.
    • $s_{2n-1}$ is decreasing and bounded below by $s_2$, so $\displaystyle \lim_{n \to \infty} s_{2n-1}$ exists.
  • And we know $\displaystyle \lim_{n \to \infty}(s_{2n} - s_{2n-1}) = \lim_{n \to \infty}(-b_{2n})$, and our conditions already says $\displaystyle \lim_{n \to \infty}(b_{n}) = 0$
  • So, $\displaystyle \lim_{n \to \infty}(s_{2n}) - \lim_{n \to \infty}(s_{2n-1}) = 0$, then we can say $\displaystyle \lim_{n \to \infty}(s_{n})$ also exits and $\displaystyle = \lim_{n \to \infty}(s_{2n}) = \lim_{n \to \infty}(s_{2n-1}) = s$.
  • And also, we know, alternating series, is between neighboring partial sums: $s_{2n-1} \le s \le s_{2n}$

Examples #

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

Limit Comparison Test #

Theorem #

  • If $a_n \ge 0$, $b_n \ge 0$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = L > 0$,
  • Then, $\sum b_n$ converges, if and only if $\sum a_n$ converges.

One of the Practice Quiz #

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

    • From the formula for geometric series shows that $$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if } \lvert y \rvert \lt 1.$$
    • And we know $x = \frac{1}{2} < 1$, so we can plug in $-x^2$ for $y$, we get that $$\begin{aligned} \frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\ &= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\ &= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots \end{aligned}$$
    • So we have $$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$
    • Then $$\begin{aligned} \arctan(x) &= \int\left(\frac{d}{dx}\arctan (x)\right),dx \\ &= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right),dx\\ &= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right),dx\\ &= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n},dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n},dx\right)\\ &= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1} }{2n+1}\right) + C\\ &= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11} }{11} +\cdots\right). \end{aligned}$$
    • Evaluating at $x = 0$ gives $0 = \arctan(0) = C$, so we get $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11} }{11} + \cdots,\qquad\text{if }|x|\lt 1.$$
    • $\frac{x^5}{5} < \frac{1}{33}$, so the approximate value of $\arctan(\frac{1}{2}) = \frac{1}{2} - \frac{1}{24} = \frac{11}{24}$

  • Note we are assuming $|x| < 1$, but arctangent is defined for all real numbers. The series we have here is $$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1} }{2n+1}.$$

  • Using the Ratio Test, we have that $$\begin{aligned} \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3} }{2n+3}\quad}{\frac{|x|^{2n+1} }{2n+1} }\\ &= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3} }{(2n+3)|x|^{2n+1} }\\ &= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\ &= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\ &= |x|^2. \end{aligned}$$

  • By the Ratio Test, the series converges absolutely if $|x|^2 < 1$ (that is, if $|x| < 1$) and diverges if $|x| > 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is 1, and the equality is valid for $x \in [-1,1]$ only.

  • This answer is copied from https://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots.

Rearrangement Theorem #

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

Example #

  • $\sum \frac{(-1)^{n+1} }{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \cdots$
  • Take the even terms, we get $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}$, which diverges.
  • Take the odd terms, we get $\sum \frac{1}{2n-1} > \sum \frac{1}{2n-1}$, which diverges too.
  • So whichever we picked, can easily get a number L as large as we want to.
Calendar July 27, 2024
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