## Literature.

G. M. Fichtenholz – Integral and differential calculus, vol. 2, PWN Warsaw 1999 (originally in Russian)

# Courses in Math - Series - Analytic Examples

This page helps you learn analytic methods how to determine series convergence or series divergence. One can become familiar with elementary theorems of theory of series convergence.

## Series of constant terms

Let us consider an infinity sequence of numbers

a1, a2, a3, …, an, …

the sum created from the numbers

a1 + a2 + a3 + … + an + …

is called infinity series, where the numbers are called series terms. Instead of the sum above one can use the sum symbol

 ∞ ∑ an (*) n = 1

where the n index varies from 1 to . For simplicity, further we will refer to this expression by the asterisk symbol: *. Let us add subsequent terms of a series

A1 = a1, A2 = a1 + a2, A3 = a1 + a2 + a3, …, An = a1 + a2 + a3 + … + an + …

The sums are called partial sums. For each series exists a sequence of partial sums {An}. A finite or infinite limit A of a sequence of partial sums An of the series sum, when n → ∞

A = lim An

is called the sum of a series, we note

 ∞ A = a1 + a2 + a3 + … + an + … = ∑ an n = 1

If a series has a finite limit then the series is called convergent. Otherwise i. e. the series limit is infinite then the series is called divergent.

## Main Theorems

### Theorem 1

If from the sum series denoted by (*) we remove m-th first terms then we get the series

 ∞ am + 1 + am + 2 + am + 3 + … + am + k + … = ∑ an n = m + 1

that is called the m-th rest of the series. If the series is convergent then each of its rest is also convergent. And oppositely, when one of the rest is convergent then the whole series is convergent too.

Proof

Let us set m and the k-th rest sum is A'k

A'k = am + 1 + am + 2 + am + 3 + … + am + k

Then one have

A'k = Am + k - Am.

When the series is convergent then An → A. If k → ∞ then the finite limit of the sequence {A'k} exists and has the form

A' = A - Am

it gives the convergence of the sum series denoted by (*).
And oppositely, when the sum series denoted by (*) is convergent i. e. A'k → A' then when k = n - m (for n > m)

An = Am + A'n - m.

when n → ∞ then

A = A' + Am

it gives that the sum is convergent.

### Theorem 2

If the series is convergent then the sum αm of its m-th rest tends to zero when m grows to ∞.

### Theorem 3

If each of sum of the convergent series is multiplied by the constant value c the its convergence is kept and its sum is multiplied by the c value.

### Theorem 4

Let us consider two convergent series

A = a1 + a2 + a3 + … + an + …

and

B = b1 + b2 + b3 + … + bn + …

One can add or subtract term by term to get the following series:

(a1 ± b1) + (a2 ± b2) + (a3 ± b3) + … + (an ± bn) + …

that is convergent too and its sum equals (A ± B).

### Theorem 5

The general term an of a convergent series tends to zero.

Proof

If An and An-1 has the finite limit A then an = An - An-1 → 0.

## Convergence of series of positive terms

Let us take a series

 ∞ ∑ an = a1 + a2 + a3 + … + an + … n = 1

that has only positive terms i. e. an ≥ 0 (n = 1,2 , 3, …). Then one get

An + 1 = An + an + 1 ≥ An,

i. e. the sequence {An} is a growing sequence. It leads to the following theorem:

### Theorem 1

The positive series (A) has always a sum. This sum is finite (i. e. the series is convergent) if partial sums of the series are limited from its upper side and is infinite (i. e. the series is divergent) in the opposite case.

This theorem very rarely allows to decide if a series is convergent or divergent. One of the examples is the divergent harmonic series:

 ∞ ∑ 1/n = 1 + 1/2 + … + 1/n + … n = 1

### Theorem 2

Le us consider two positive series

 ∞ ∑ an = a1 + a2 + a3 + … + an + … n = 1
and
 ∞ ∑ bn = b1 + b2 + b3 + … + bn + … n = 1

If starting from some point (e. g. n > N) the inequality is obeyed an ≤ bn, then the convergence of the second series (B) implies the convergence of the first one (A) or the in-convergence of the series A implies the in-convergence of the series B.

### Theorem 3

If the following limit exist

lim (an / bn) = K and (0 ≤ K ≤+∞)

and since K < +∞ then if the series B is convergent it implies that the series A is convergent too. And since K > 0, in-convergence of the A series leads to in-convergence of the B series.

### Theorem 4

If starting from some point (e. g. n > N) the below inequality is fulfilled

an+1 / an ≤ bn+1 / bn

then if the series B is convergent it implies that the series A is convergent too. And the in-convergence of the A series leads to the in-convergence of the B series.

## Cauchy Condition

Let us create for the A series a sequence

cn = (an)1/n.

If for n that is great enough the following inequality is obeyed

cn ≤ q,

where q is a number less than 1, then this series is convergent. If starting from some point is

cn ≥ 1,

then the series is divergent.

### Cauchy Condition - Second Formulation

Let us assume that a sequence {cn} has a limit (finite or infinite)

lim cn = C when n → ∞

If C < 1 then the series is convergent and if C > 1 then the series is divergent.

### D'Alembert Condition

Let us consider for the A series a sequence

dn = an+1/an.

If for n that are great enough the following inequality is obeyed

dn ≤ q,

where q is a constant number less than 1, then this series is convergent. If starting from some point is

dn > 1,

then the series is divergent.

### D'Alembert Condition - Second Formulation

Let us assume that the sequence {dn} has a limit (finite or infinite)

lim dn = D when n → ∞.

If D < 1 then the series is convergent and if D > 1 then the series is divergent.

### Raab Condition

Let us consider for the A series a sequence

rn = n(an/an + 1 - 1).

If for n that are great enough the following inequality is obeyed

rn ≥ r,

where r is a number greater than 1, then this series is convergent. If starting from some point is

rn < 1,

then the series is divergent.