Problem

Determine whether the sequence converges or diver…

02:46

Problem 44 Easy Difficulty

Determine whether the sequence converges or diverges. If it converges, find the limit.
$ a_n = \sqrt [n]{2^{1 + 3n}} $

Answer

The sequence converges to 8

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 11

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Video Transcript

So I think the easiest way to approach this problem is by rewriting this A in term. So this in that we see happening over there, That just means that all this stuff is being taken to the one over in power. So the one over n And then as long as you know how exponents work, then you know that this one over in is going to distribute to all of this stuff. So this turns into one times one over in plus three in times one over in, which is just three. So now if we do limit as n goes to infinity of a n mhm, we get limit as n goes to infinity of two to the one over N Plus three and to to the X is a continuous function, which means that you're allowed to pull the limit inside of the function. So this is the same thing as two to the power of limit as n goes to infinity of one over n plus three As n goes to infinity, one over N goes to zero and we get two to the three, which is eight. So our sequence converges and it converges to eight. Yeah,

Gabriel R.

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

Problem