Higher Mathematics. Part 3

ISBN 978–966–598–612–6

© Denisiuk V. P., Grishina L. I.,

Karupu O. V. and fo on, 2009

ISBN 978–966–598–613–3 (Part 3)

© NAU, 2009

∞

a1 + a2 + a3 + ...+ an + ... = ∑ an

(1.1)

n=1

n

Sn = a1 + a2 + a3 + ... + an = ∑ ai .

(1.2)

i=1

Consider the sequence {Sn } of partial sums of the series (1)

S1 = a1 ,

S2

= a1 + a2 ,

.............

Sn

= a1 + a2 + a3 + ... + an ,... .

Definition. If the sequence

{Sn }

has a finite limit,

lim Sn = S,

i. e.,

{Sn }

∞

converges to S, then the limit is called the sum of the series ∑ an

and the se-

∞

n=1

= S. If there is no

lim Sn , i.e., {Sn }

ries is said to be convergent: ∑ an

does

n=1

∞

not converge, then the series

∑ an

is said to be divergent and does not have

any sum.

n=1

Example. Examine convergence of the series

∞

(1.3)

1+1+1+...+1+... = ∑1n

n=1

We consider the nth partial sum of the series

Sn =1+1+1+...+1= n.

Then lim Sn = lim n = ∞.

n→∞

n→∞

1−1+1−1+...+

∞

(1.4)

(−1)n−1 +... = ∑(−1)n−1

n=1

We consider the sequence of partial sums {Sn}

of the series (1.4)

S1 = 1,

S2 = 0,

S3 = 1, S4 = 0, ..., S2n−1 = 1, S2n = 0

and understand that lim Sn

does not exist.

n→∞

Then the given series diverges.

Consider a series known as a geometric series with a ratio q

∞

−1, (a ≠ 0).

a + aq + aq2 + aq3 + ... + aqn−1 + ... = ∑ aqn

n=1

The sum of n terms is

Sn

= a + aq + aq2 + aq3 + ... + aqn−1 =

=

a − aqn

=

a

−

aqn

,

(q ≠ 1).

1− q

1

− q

1− q

If

q

< 1 , then lim qn

= 0 , and so

n→∞

a

aqn

a

lim Sn = lim

−

=

,

− q

1− q

1

− q

n→∞

n→∞

1

i.e., the series converges and its sum is

a

, or

1

− q

∑ aqn−1

=

a .

∞

n=1

1− q

then lim qn

If

q

> 1,

= ∞ and hence lim Sn

= ∞,

i. e., the series diverges.

n→∞

n→∞

At

q = −1 we obtain the divergent series a − a + a − a + ... (a ≠ 0).

Its par-

a for odd n,

tial sum is Sn =

q =1

0 for even n.

a + a + a + a + ... ,

for which Sn = an, and

At

we will have the series

hence lim Sn = lim an = ∞,

i. e., the series diverges.

n→∞

n→∞

6

The geometric series

∞

−1,

(a ≠ 0)

∑ aqn

n=1

converges for

q

< 1 and diverges for

q

≥ 1 .

Definition. Consider the number series

(1.5)

1+ 1 + 1 + ... +

1 + ... = ∑ 1,

∞

2

3

n

n=1

n

known as the harmonic series.

We will prove that the series is divergent. We remember that

1 n

1+

> e , for any n N.

n

Passing to the natural logarithm we obtain

1

n ln 1+

< 1

n + 1

1

1

n

> ln(n

+ 1) − ln(n).

or

ln

<

and let’s write

n

n

n

Hence

if

n = 1 then

1 > ln 2 − ln1,

if

n = 2 then

1

> ln 3 − ln 2,

2

if

n = 3 then

1

> ln 4 − ln 3,

3

. . . . . . . . . . . . . . . . . . . . . . . . . . .

if

n

then

1

> ln(n + 1) − ln(n).

n

Let’s find the sum of the right and left hand parts

1+

1

+

1

+ ... +

1

> (ln 2 − ln1) + (ln 3 − ln 2) + (ln 4 − ln 3) + ...

3

2

n

... + (ln(n + 1) − ln(n)) = ln(n + 1).

So we have Sn

> ln(n + 1) .

Passing to the limit we obtain lim Sn

= ∞, then the harmonic series is diver-

gent.

n→∞

7

It follows that if there exists lim Sn , then there is lim Sn− k (k = const) and,

n→∞

n→∞

conversely, if there is lim Sn− k , then there exists lim Sn .

n→∞

n→∞

Remark. The resultant series a

ak +1 + ak + 2 + ... has the sum S = S − σk , S

being the sum of the original series.

∞

∞

∑ λan = λ ∑ an .

(1.6)

n=1

n=1

∞

∞

Proof. We write partial sums for the series ∑ an and

∑ λan :

n=1

n=1

Sn

= a1 + a2 + ... + an and

σn = λa1 + λa2 + ... + λan .

It is obvious that σn

= λSn .

As stated, the series

∞

Sn , then from the

∑ an converges, i. e., there exists lim

last equality there is

n=1

n→∞

lim σn , so that

n→∞

lim σn

= lim λSn

= λ lim Sn , i. e.,

n→∞

n→∞

n→∞

∞

∞

∑ λan = λ ∑ an .

n=1

n=1

8

∞

∞

Theorem 3

If the series ∑ an

and ∑ bn

converge, then their sum and dif-

n=1

n=1

∞

∞

ference, i. e., ∑ (an + bn )

and ∑ (an − bn ), converge, and

n=1

n=1

∞

∞

∞

∑ (an ± bn ) = ∑ an ± ∑ bn .

n=1

n=1

n=1

Proof. Let Sn

= a1 + a2 + ... + an ,

σn = b1 + b2 + ...+ bn

and

Cn = (a1 + b1 ) + (a2 + b2 ) + ... + (an + bn ),

∞

∞

∞

be partial sums of the series ∑ an ,

∑ bn and

∑ (an + bn ),

respectively.

Clearly, σn + Sn = Cn .

n=1

n=1

n=1

∞

∞

As by the statement of the theorem the series ∑ an and ∑ bn , converge, i.e.,

there exist lim Sn

n=1

n=1

and lim σn , it follows from the last equality, which holds for

n→∞

n→∞

all n, that there exists lim Cn , and

n→∞

(Sn + σn ) = lim Sn + lim σn ,

lim Cn

= lim

n→∞

n→∞

n→∞

n→∞

which is equivalent to

∞

∞

∞

∑ (an

+ bn ) = ∑ an + ∑ bn .

n=1

n=1

n=1

∞

In the same manner, we can prove the convergence of ∑(an − bn ) .

n=1

Let introduce the concept of the remainder of a series. If we discard the first

a1 + a2 + ... + an + an+1 +

∞

n terms in the convergent series

+ an+ 2 + ... = ∑ an we

n=1

∞

will obtain the convergent series

an+1 + an+ 2 + ... ... + an+ k + ... = ∑ an+ k which is

k =1

Rn

∞

called the n-th remainder of the series and denoted by

= ∑ an+ k for a fixed n.

k =1