Homework ( Solutions )
1. ¥
  å
  n = 1
  2n+5
  (n+1)²
  .
2. ¥
  å
  n = 1
  1
  2ⁿ-1
  .
3. ¥
  å
  n = 1
  cosn
  n
  .
- Series of nonnegative terms
- Alternating series..absolute and conditional convergence
- Power series
Series of nonnegative terms
- Does it converge, if so what does it sum to?
- Partial sums of these form nondecreasing sequences, and if these are bounded then we have convergence.
- Show
  
  å
  n
  1
  n²
  
  converges. Can view the sequence as a function and the series as the area of rectangles, which is bounded by 1+ò₀^¥ 1/x²dx = 2. Since bounded know that converges so done.
- Integral test: Let a_n be positive terms in a sequence, where a_n = f(n) and f is a continuous, positive, decreasing function of x for all x > N. (N positive integer) Then the series
  
  ¥
  å
  n = N
  
  and in the integral
  
  ó
  õ ¥
  
  N
  f(x)dx
  
  both converge or diverge.
- Harmonic series and the p-series
  
  ¥
  å
  n = 1
  1
  n^p
  
  converges if p > 1 and diverges if p £ 1.
- p-series with p = 1 are called harmonic series. Tests show that if increase p to 1.0000000001 the series converges. Approaches infinity incredibly slowly. For example need to add 178,482,301 terms to reach 20.
- Comparision test
  - Direct comparision test:Let åa_n be a series of nonnegative terms.
    1. åa_n converges if there is a convergent series åc_n with a_n £ c_n for all n > N.
    2. å_n a_n diverges if there is a divergent series of nonnegative terms åd_n with a_n ³ d_n for all n > N.
  - Limit comparision test
    
    lim
    a_n
    b_n
    
    If this is 0 and b_n converges then a_n converges. If finite number then a_n and b_n both converge or diverge. If ¥ and b_n diverges then a_n diverges too.
- Ratio test
  
  lim
  n® ¥
  a_n+1
  a_n
  = p.
  
  If p < 1 then converges, if p > 1 then diverges. If p = 1 then this test is inconclusive.
- nth-root test:
  
  lim
  n® ¥
  (a_n)^1/n = p.
  
  Converges if p < 1, diverges if p > 1. Inconclusive if p = 1.
Alternating series
- A series in which terms are alternating sign is...an alternating series!
- Series is å(-1)ⁿ⁺¹ u_n.
- Converges if satisfy all of following:
  1. u_n's positive.
  2. u_n ³ u_n+1.
  3. u_n ® 0.
- If series satisfies all of these then the truncation for the nth partial sum is less than u_n+1 and has the same sign as the unused term.
- å(-1)ⁿ (1/2ⁿ)=1-1/2+1/4-1/8+... this is equal to
  
  1
  1-(-1/2)
  = 2/3.
  
  If add upto 1/128 then know that error is less than 1/256..and positive. Actual error is 0.0026 while 1/256 is 0.0039.
- Absolute convergence:A series converges absolutely if the series of absolute value values a_n converges.
- A series converges conditionally if it converges but it does not absolutely.
- Absolute convergence is important since if a series converges absolutely then it converges and we have better tests for positive series.
- Alternating p-series: Converges absolutely if p > 1, conditionally if 1 ³ p > 0.
- rearrangement..and absolutely convergent series converges to the same result even if rearranged(won't go into that). An alternating harmonic series can be rearranged to diverge or to reach any preassigned sum.
Power series
- å
  c_n xⁿ
  
  is a power series centred at x = 0.
  
  å
  c_n(x-a)ⁿ
  
  is centred at x = a.
- Convergence test for power series: three posibilities
  1. There is a positive number R such that the series diverges for | x-a | > R but converges for | x-a | < R. The series may or may not converge at x = a+R or x = a-R.
  2. The series converges for all x.
  3. The series converges at x = a and diverges elsewhere.
- The number R is called the radius of convergence, and the set of x for which the series converges is called the interval of convergence.
- A few points about power series then to Taylor, on Tue. Application of power series to differential equations and fourier on friday, then revision and sample papers.
  Next lecture

File translated from T_EX by T_TH, version 2.70.
On 3 May 2002, 13:15.