13-16 матан

13.Cauchy’s convergence criterion.

∞

A series ∑ An converges iff for every Ԑ >0 there is nԐ<N and any p€ N

n=1

|Sn+p - Sn|<Ԑ |an+1 +an+2 +..+an+p|<Ԑ (1) ∞

Proof: In terms of the partial sums {Sn} at ∑ an, Sn+p – Sn= an+1 +an+2 +..+an+p

n=1

Therefore, (1) can be written as |Sn+p - Sn|<Ԑ. Since ∑ an converges if and only if {Sn} converges. 13.Cauchy’s convergence criterion for sequence implies the conclusion.

14. Comparison test 1,2,3.

Comparison test 1.

∞ ∞

Let ∑ an and ∑ Bn be series with nonnegative terms and suppose that an ≤ Bn(4)

n=1 n=1

1. If the “bigger” series ∑ Bn converges, then the “smaller” series ∑an also converges.

2. If the “smaller” series ∑an diverges, then the “bigger” series ∑Bn also diverges.

Proof: (a) if An= ak+ak+1..an and Bn=bk+bk+1..bn, n ≥ k, then the from (4) An ≤Bn(5)

If ∑Bn  ,then {Bn} is bounded above and (5) implies that {An} is also; therefore ∑an.

On the other hand, if ∑an diverges, then the sequence {An} is unbounded above and (5) implies that {Bn} also; therefore ∑Bn diverges.

Comp. test 2

Suppose that an≥ and Bn>0 and lim an/Bn =L,where L.>0. Then ∑an and ∑ bn converges or diverges together.

Comp. test 3

∞

∑ 1/n^p,  if p>1

n=1 diverges if p≤1

15. The D’alemberts test. THE RAABE’S TEST. THE GAUSS’S TEST.

The D’alemberts test

Let ∑an be a series with positive terms and suppose that lim=An+1/An=p

n->∞

1. If p<1, the series 

2. If p >1, the series diverges

3. If p=1, the series may converge and diverge, so that another test must be tried.

Proof: __ __

Suppose that an>0 for n≥k. Then ∑an if lim An+1/an>1 because if lim An+1/an such that

n->∞ n->∞

o<r<1 and An+1/an< r^(n+1)/ r^n.

Since ∑r^n / we use comp.test 1 and the series ∑an . If lim An+1/an>1 there is a number r

---

Such that r>1 and An+1/an>r for n sufficiently large. This can be rewritten as An+1/an > r^(n+1)/r^n. since ∑r^n diverges. Comp test 1 implies that ∑ bn diverges.

Then lim An+1/an < 1, ∑an 

limAn+1/an>1, ∑an diverges.

THE RAABE’S TEST

Suppose that an>0 for large n.

Let lim n( an/An+1 – 1)= p , if p>1, the series 

n->∞ if p<1 , diverges

if p=1, ∑an-?

THE GAUSS’S TEST.

Suppose that an>0, for any n€N

Let an/An+1= ϒ(альфа)+ µ/n +0(c 2мя точками сверху) (1/n)

Then if ϒ>1 => ∑an 

If ϒ<1 => ∑an diverges

If ϒ=1

µ=1 => ∑ an 

if ϒ=1, µ<1 => ∑ an diverges.

16. Cauchy’s root test. Absolute and conditional convergence. Leibnitz’s test. Derichlet’s test. Abel’s test.

Cauchy’s root test.

∞

Let ∑ be a series with positive terms and suppose that lim ak под корнем в степени к= L

n=1 k∞

1. If L<1, the series 

2. If L>1, the series diverges

3. If l=1, may converges or diverges, so that another test must be tried.

Absolute and conditional convergence.

∞

A series ∑ ak= a1+a2+…+ak+… is said to convergence absolutely, if the series of absolute

k=1 ∞

values ∑ |ak|= |a1|+|a2|+…+|ak|+… converges too.

n=1

If the series that converges, but diverges absolutely is said to converges conditionally.

Leibnitz’s test. ∞

∑ (-1)^n an (1)  if

n=1

1. (1) is alternating series

2. An ↓ (monotonically decreasing)

3. Lim an=0

n->1

Derichlet’s test.

∞

∑ an bn  if

n=1 ∞

1. | ∑ an| ≤ k there is k>0, partial sum is bounded

n=1

∞

2. {bn} ↓

N=1

Abel’s test.

∞

∑ an bn  if

n=1

∞

1. ∑ an 

n=1

∞

2. {bn} monotonically ↓↑

n=1

3. |bn| ≤ c , there is c>0