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# Stolz-Cesaro Theorem

## 1.

Calculus++ Light## 2.

Question 4. Find the limit of the sequence2, 2 2 , 2 2 2 , 2 2 2 2 ,

Solution. Our sequence can be written down

as follows

1

2

2 ,2

1 ( 1 )2

2

2

,2

1 ( 1 ) 2 ( 1 )3

2

2

2

,2

1 ( 1 ) 2 ( 1 )3 ( 1 ) 4

2

2

2

2

,

Therefore the nth term of the sequence is given

1 ( 1 ) 2 ( 1 ) 3 ( 1 ) 4 ( 1 ) n

2

by

an 2 2 2 2 2

.

Using the formula for the sum of a geometric

series we obtain

1

1 n 1

(2)

( ) ( ) ( )

1

1

2

1 )n

n

lim

1

(

1

2

1 ( 2 )

n

2.

lim an lim 2 2

1

2

n

1 2

2

1 3

2

n

1 n

2

2

1 ( ) .

1 n

2

## 3. Stolz-Cesaro Theorem

Let an and bn be two sequences of real numbers.Assume that:

I. bn as n ,

II. bn is increasing for sufficiently large n,

an 1 an

III. lim

L.

n b

n 1 bn

an

Then lim

L.

n b

n

an

Question 1. an ln n, bn n, lim

?

n b

n

Solution. The conditions I and II of the StolzCesaro theorem are satisfied.

## 4.

an 1 anln( n 1) ln n

lim

lim

n b

n

b

n 1 n

n 1

n

n 1

1

lim ln 1

lim ln

n

n

n

n

1

ln lim 1 ln 1 0.

n n

Thus, the Stolz-Cesaro Theorem tells us that

an

ln n

lim

lim

0.

n b

n n

n

2

n

To find the limit lim n

n 2

either apply the Stolz-Cesaro Theorem

twice,

## 5.

or write it down as a productlim

n

n

2

n

n

2

n

,

apply the Stolz-Cesaro Theorem to lim

n

and then use the product rule

lim xn yn lim xn lim yn .

n

n

n

n

2

n

## 6.

Answers to Questions from Light #1:Sequences and Limits

1

1

1

Question 0: an

n(n 1) n n 1

17

Question 1:

a650 Question 2: nmin 11

20

Question 4: 10

3

Question 5: x log 2

2018

1 2

A 3, B 2, C 2018, D 2

## 7.

Calculus++Also known as

Hysterical Calculus

## 8.

Question 1a. Find the following limit1

1

1

1

1

2

3

n

lim

.

n

n

Solution. Use the Stolz-Cesaro theorem.

In this case bn n .

The sequence bn is infinitely large and

increasing. Hence, the conditions I and II of

the Stolz-Cesaro theorem are satisfied.

1

an

1

1

an 1

1

1

2

1

2

1

1

an 1 an

n

n 1

1

1

n

n 1

## 9.

an 1 an1

1

bn 1 bn

n 1 n 1 n

1

n 1 n

1

n 1 n 1 n n 1 n

1

n 1 n

1

n 1

an 1 an

n

2.

lim 1

Hence lim

n b

n

b

n

1

n 1

n

The Stolz-Cesaro Theorem tells us that

1

1

1

1

an

1

2

3

n

2.

lim

lim

n

n b

n

n

## 10. Cauchy Criterion

A sequence xn, n = 1,2,3,… is called afundamental sequence (or Cauchy

sequence) if for any 0 we can find a

number N such that, for any n > N and any

m > 0:

x x .

n m

n

Theorem (Cauchy Criterion). A sequence xn,

n = 1,2,3,…, converges if and only if it is a

Cauchy sequence.

## 11.

Definition (of non-fundamental sequences).A sequence xn, n = 1,2,3,… is not a Cauchy

sequence if we can find 0 such that, for

any number N, we can find n > N and m > 0,

such that xn m xn .

## 12.

Question 3. The sequence –1, +1, –1, +1,… isnot a Cauchy sequence (and, hence, it

diverges).

Solution. Let 1, and let N be any natural

number. Take n = 2N + 1, m = 1.

Since n is odd and n + m is even, we have

xn= –1 and xn+m = +1.

Hence xn m xn 1 ( 1) 2 .

Therefore, the sequence

{xn} = –1, +1, –1, +1, …

is not a Cauchy sequence.

## 13.

Question 4. Use the Cauchy criterion to showthat the sequence

n

1

xn , n 1,2,3, ,

k 1 k

diverges.

Solution: According to the Cauchy criterion it

is sufficient to show that {xn} is not a

fundamental sequence:

0, N , n N , m 0 : xn m xn .

We have

n m

1

xn m xn

k n 1 k

n m

m

1

.

n m

k n 1 n m

## 14.

Choosing m = n we obtainm

n

1

xn m xn

.

n m n n 2

Thus, 0 (for instance, 14 ), N

n N (for instance, n N 1),

1

(we

set

m

=

n):

x

x

m 0

n m

n

4.

1

2

Therefore, our sequence {xn} is not

fundamental, and the Cauchy criterion tells

us that {xn} diverges.

## 15.

Question 5. Use the Cauchy criterion to shown

1

that the sequence xn 2 , n 1,2, ,

k 1 k

converges.

Solution: It is sufficient to show that the

sequence xn is fundamental:

0, N , n N , m 0 : xn m xn .

n m

n m

1

1

We have xn m xn 2

k n 1 k

k n 1 ( k 1) k

n m

1 1

1

1

1

1

k n n 1 n 1 n 2

k n 1 k 1

1

1

1

1

n m 1 n m

n 2 n 3

## 16.

m1

1

1

.

n n m n ( n m) n

1

Thus xn m xn .

n

1

Therefore 0, N , we set N

,

n N , m 0 :

1 1

1

xn m xn 1 .

n N

Thus, the sequence xn is fundamental, and

therefore it converges to some limit L.

In fact, L

2

6

.

## 17. Picture of the Week

## 18.

Question 8. Draw the curve defined by the1

equation lim

n n

n

| x| | y|

x y

.

2

2

n

2

Solution. We already know that

lim | x | | y | max(| x |, | y |).

n

1

Therefore lim

n

n

n

| x| | y|

n

n

n

| x| | y|

| x || y |

| x || y |

lim

n

n

n n

max(| x |, | y |)

| x| | y|

min(| x |, | y |).

n

n

n

## 19.

Thus, we have to draw the curve defined bythe equation

2

2

x y

min(| x |, | y |)

.

2

## 20.

Let us look at the xy – plane:y x

min(| x |, | y |) | x |

y x

y

1

min (| x |, | y |) | y |

min(| x |, | y |) | y |

x

1

1

1

min(| x |, | y |) | x |

## 21.

x yThe curve defined by the equation y

2

2

2

is the circle with the radius 1, centred at the

point (0,1).

2

2

Indeed,

x y 2y 0

x y 2 y 1 1

2

2

x ( y 1) 1.

2

2

x y

The curve defined by the equation x

2

2

is the circle with the radius 1, centred at the

point (1,0):

2

2

( x 1) y 1.

2

## 22.

The equations of our curve.( x 1) y 1 ( x 1) y 1

y

y x

1

y x

2

2

2

x ( y 1) 1

x

2

x ( y 1) 1

2

2

2

1

2

2

x ( y 1) 1

2

1

2

2

x ( y 1) 1

( x 1) y 11

2

2

( x 1) y 1

2

2

## 23.

lim1

n n

n

| x| | y|

x y

.

y 2

2

n

1

2

x

1

1

1

The picture of the week.

## 24.

Answers to Questions from Seminar 1.Question 2: 13 .

n

n

5

1

Question 3: an 6 2 3 ( 1)

a100 a99 5 2

Question 8b:

98

lim sin n n 1.

2

n

2

2 n

Question 8c: lim sin

0.

n

3n 1

1, for 0 x 1;

Question 9: y x, for 1 x 2;

1 x 2 , for 2 x.

2

n