Week 2 - Series

1. Definition

  • In general, if we try to add the terms of an infinite sequence {an}n=1\{a_n\}_{n=1}^{\infty} we get an expression of the form a1+a2+a3++an+a_1+a_2+a_3+\cdots+a_n+\cdots
  • which is called an infinite series (or just a series) and is denoted, for short, by the symbol n=1an or an\sum_{n=1}^{\infty} a_n \ \text{or}\ \sum{a_n}

1.1. Convergent and Divergent

  • Given a series n=1an=a1+a2+a3+\sum_{n=1}^{\infty}a_n = a_1 + a_2 + a_3 + \cdots, let sns_n denote its nth partial sum: sn=i=0nai=a1+a2+a3++ans_n = \sum_{i=0}^n a_i = a_1 + a_2 + a_3 + \cdots + a_n
  • If the sequence {sn}\{s_n\} is convergent and limnsn=s\lim_{n \to \infty} s_n = s exists as a real number, then the series an\sum a_n is called convergent and we write a1+a2+a3++an+=s or n=1an=sa_1 + a_2 + a_3 + \cdots + a_n + \cdots = s \ \text{or}\ \sum_{n=1}^{\infty} a_n = s
  • The number s is called the sum of the series. If the sequence {sn}\{s_n\} is divergent, then the series is called divergent.

2. Geometric Series

  • a+ar+ar2+ar3++arn1+=n=1arn1, a0a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} + \cdots = \sum_{n=1}^{\infty} ar^{n-1}, \ a \ne 0
  • Each term is obtained from the preceding one by multiplying it by the common ratio r.
  • If r=1r = 1, then sn=a+a+a++a=na±s_n = a + a + a + \cdots + a = na \to \pm \infty. Since limnsn\lim_{n \to \infty} s_n doesn't exist, the geometric series diverges in this case.
  • If r1r \ne 1, we have sn=a+ar+ar2+ar3++arn1rsn=ar+ar2+ar3++arn\begin{aligned} s_n &= a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} \\ rs_n &= ar + ar^2 + ar^3 + \cdots + ar^{n} \end{aligned}
    • Subtracting these equations, we get snrsn=aarns_n - rs_n = a - ar^n
    • sn=a(1rn)1rs_n = \frac{a(1-r^n)}{1-r}
  • If 1r1-1 \le r \le 1, then rn0, as nr^n \to 0,\ \text{as}\ n \to \infty, so limnsn=limna(1rn)1r=a1ra1rlimnrn=a1r\lim_{n \to \infty}s_n = \lim_{n \to \infty}\frac{a(1-r^n)}{1-r} = \frac{a}{1-r} - \frac{a}{1-r}\lim_{n \to \infty}r^n = \frac{a}{1-r}
  • Summarize the results The geometric series n=1arn1=a+ar+ar2+\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \cdots is convergent if r<1|r| < 1 and its sum is n=1arn1=a1r    r<1\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}\ \ \ \ |r| < 1 If r1|r| \ge 1, the geometric series is divergent.
  • Another way to get the conclusion (1r1-1 \le r \le 1):
    • This figure provides a geometric demonstration of the result in the example. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles, sa=aaar , so s=a1r\frac{s}{a} = \frac{a}{a- ar}\ \text{, so}\ s= \frac{a}{1-r}
  • Example: 0.9ˉ0.\bar{9}
    • 0.9ˉ=n=1910n=9n=110n=9n=1110n=91/1011/10=919=1\begin{aligned} 0.\bar{9} &= \sum_{n=1}^{\infty}9 \cdot 10^{-n} \\ &= 9 \sum_{n=1}^{\infty}10^{-n} = 9 \sum_{n=1}^{\infty}\frac{1}{10^n} \\ &= 9 \cdot \frac{1/10}{1- 1/10} = 9 \cdot \frac{1}{9} = 1 \end{aligned}
      • Notice the formula is arn1ar^{n-1}, so here =91/1011/10= 9 \cdot \frac{1/10}{1- 1/10}

3. Telescoping Series

  • A telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation.
  • For example, the series n=11n(n+1)\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}
    • (the series of reciprocals of pronic numbers) simplifies as n=11n(n+1)=n=1(1n1n+1)=limNn=1N(1n1n+1)=limN[(112)+(1213)++(1N1N+1)]=limN[1+(12+12)+(13+13)++(1N+1N)1N+1]=limN[11N+1]=1.{\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}&{}=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\sum _{n=1}^{N}\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\left\lbrack {\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1-{\frac {1}{N+1}}}\right\rbrack =1.\end{aligned}}
  • From Wikipedia

4. Harmonic Series

  • n=11n=1+12+13+14+\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots is divergent.
  • Solution: For this particular series it’s convenient to consider the partial sums s2,s4,s8,s16,s32,s_2, s_4, s_8, s_{16}, s_{32}, \ldots and show that they become large.
    • Similarly, s32>1+52,s64>1+62s_{32} > 1 + \frac{5}{2}, s_{64} > 1 + \frac{6}{2}, and in general s2n>1+n2s_{2^n} > 1 + \frac{n}{2}

5. Theorem

  • If the series n=1an\displaystyle\sum_{n=1}^{\infty} a_n is convergent, then limnan=0\displaystyle\lim_{n \to \infty} a_n = 0.
  • If limnan\displaystyle\lim_{n \to \infty} a_n does not exist or if limnan0\displaystyle\lim_{n \to \infty} a_n \ne 0, then the series n=1an\displaystyle\sum_{n=1}^{\infty}a_n is divergent.
  • If an\sum a_n and bn\sum b_n are convergent series, then so are the series can\sum c a_n (where c is a constant), (an+bn)\sum(a_n + b_n), and (anbn)\sum(a_n - b_n), and
    • n=1can=cn=1an\displaystyle\sum_{n=1}^{\infty} c a_n = c \sum_{n=1}^{\infty} a_n
    • n=1(an+bn)=n=1an+n=1bn\displaystyle\sum_{n=1}^{\infty} (a_n + b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n
    • n=1(anbn)=n=1ann=1bn\displaystyle\sum_{n=1}^{\infty} (a_n - b_n) = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n

6. The Comparison Tests

  • The Comparison Test Suppose that an\sum a_n and bn\sum b_n are series with positive terms.
    • If bn\sum b_n is convergent and an<bna_n < b_n for all n, then an\sum a_n is also convergent.
    • If bn\sum b_n is divergent and an>bna_n > b_n for all n, then an\sum a_n is also divergent.
  • Example: Does n=11n2\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} converges?
    • Let's start with n = 2, which should get the same conclusion.
    • n=21n2<n=11n(n1)=n=1(1n11n)=limn11n=1\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} < \sum_{n=1}^{\infty} \frac{1}{n \cdot (n-1)} = \sum_{n=1}^{\infty} (\frac{1}{n -1} - \frac{1}{n}) = \lim_{n \to \infty} 1 - \frac{1}{n} = 1
    • 0<n=21n2<10 < \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} < 1

7. Cauchy Condensation

  • The sequence {ak}\{a_k\} decreasing and ak>0a_k > 0. The series k=1ak\displaystyle\sum_{k=1}^{\infty} a_k converges if and only if k=02ka2k\displaystyle\sum_{k=0}^{\infty} 2^k \cdot a_{2^k} converges.
  • Proven:
    • ak>0a_k > 0, so k=1ak\displaystyle\sum_{k=1}^{\infty} a_k is non-decreasing.
    • Same as the Harmonic Series part, we can get: sn=1+(a2+a3)+(a4+a5+a6+a7+a8)+sn<1+2a2+4a4+sn<k=02ka2k\begin{aligned} s_n &= 1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7 + a_8) + \cdots \\ s_n &< 1 + 2 \cdot a_2 + 4 \cdot a_4 + \cdots \\ s_n &< \sum_{k=0}^{\infty} 2^k \cdot a_{2^k} \end{aligned}
  • Same example: Does n=11n2\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} converges?
    • First way -> Use Telescoping Series:
      • n=11n2\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} converge, iff n=21n2 \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} converge.
      • 01n21n2n  n20 \le \frac{1}{n^2} \le \frac{1}{n^2 - n}\ \text{, }\ n \ge 2 => if n=21n2n\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n}, then n=21n2\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} converge.
      • n=21n2n=n=2(1n11n)\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n} = \sum_{n=2}^{\infty} (\frac{1}{n-1} - \frac{1}{n}) which is a telescoping series.
      • limnn=2N(1n11n)=limn(11N)=1\displaystyle \lim_{n \to \infty}\sum_{n=2}^{N} (\frac{1}{n-1} - \frac{1}{n}) = \lim_{n \to \infty}(1 - \frac{1}{N}) = 1
      • So n=21n2n\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n} converge => n=21n2\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} converge => n=11n2\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} converge
    • Second way -> Use Conchy Condensation:
      • The sequence {1n2}\{\frac{1}{n^2}\} decreasing and 1n2>0\frac{1}{n^2} > 0, So:
      • n=11n2<n=02n1(2n)2=n=012n=2\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} < \sum_{n=0}^{\infty} 2^n \cdot \frac{1}{(2^n)^2} = \sum_{n=0}^{\infty} \frac{1}{2^n} = 2
      • With Geometric Series n=012n=111/2=2\displaystyle\sum_{n=0}^{\infty} \frac{1}{2^n} = \frac{1}{1-1/2} = 2

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