Question

e=series of (1/n!) n=0 b=infinity

***My professor never taught us how to find upper and lower bounds of course and I believe there will be an exam question on it so help would be appreciated:D

(a) Find a lower bound for e by evaluating the first five terms of the series
(b) Show that (1/n!) < or equal to (1/2^(n-1)) for n> or equal to 1.
(c) Find an upper bound for e using part (b)

Answer 1

Well the terms in this series are strictly positive, which means that if you evaluate the first five terms, the full sum is going to be greater than what you just found, right?

Answer 2

For a, just evaluate the first five terms of the series algebraically, as @Jemurray3 said

Answer 3

well, less algebra and more arithmetic I guess

Answer 4

how would i do that? just plug in 1,2,3,4,5 for n?

Answer 5

actually, n=0,1,2,3,4

Answer 6

i know this might sound stupid but i forgot what the ! means....is it just if you have (1/5!) its going to be ((1)/(1*2*3*4*5))?

Answer 7

and oh yeah thank you!

Answer 8

yup

Answer 9

n!=n(n-1)(n-2)(n-2)...(1)

Answer 10

i am thinking that induction is the way to go for the second part, but there may be an easier way to see it

Answer 11

"n!=n(n-1)(n-2)(n-2)...(1)" .....what?

Answer 12

uhh, just ignore that I guess

Answer 13

so its 1,1,(1/2),(1/6),(1/24)

Answer 14

for part b, factor the expressions in the denominator, kinda like @satellite73 would say you could use induction; and yes, 1,1,1/2,1/6,1/24

Answer 15

what do u mean factor the expressions?

Answer 16

"Show that (1/n!) < or equal to (1/2^(n-1)) for n> or equal to 1."
take a general term in the series n, and factor the denominator for 1/n!, showing that 1/n! >1/2^(n-1)

Answer 17

i guess induction is kind of overkill, but you do have an \(n\) there
the second part asks you to show \(\frac{1}{n!}<\frac{1}{2^{n-1}}\) but this is more or less obvious, since if \(n>2\) you have
\[\frac{1}{n!}=\frac{1}{n}\times\frac{ 1}{n-1}\times \frac{1}{n-2}\times ...\frac{1}{2}<\frac{1}{2}\times \frac{1}{2}\times ...\times \frac{1}{2}\]

Answer 18

For \(n\geq 2\)
\[
\underbrace{2\cdots2}_{n-1}\leq 1\cdot 2\cdots n
\]

Answer 19

so guess what your upper bound is?

Answer 20

I never liked geometric series

Answer 21

im not sure im still confused with b

Answer 22

i dont either lol

Answer 23

lets to it for \(n=4\)

Answer 24

the (1/2)s are confusing me on what u wrote above...

Answer 25

\(\frac{1}{4!}=\frac{1}{4\times 3\times 2\times 1}=\frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}\times 1\)

Answer 26

ooooooh!

Answer 27

in this case \(n=4\) and so \(n-1=3\)

Answer 28

and it should be clear that \[\frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}<\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\] since all the denominators on the left are greater than or equal to the denominator of 2 on the right

Answer 29

generalize to show that \(\frac{1}{n!}<\frac{1}{2^{n-1}}\)

Answer 30

okay so then how do i find the upper bound?

Answer 31

not really much to generalize
the denominators on the left get larger and larger, whilst the denominator on the right is always 2

Answer 32

use part b
since \(\frac{1}{n!}<\frac{1}{2^{n-1}}\) you know
\[e=\sum_{n=1}^{\infty}\frac{1}{n!}<\sum_{n=0}^{\infty}\frac{1}{2^{n-1}}\]
and the one on the right is your upper bound

Answer 33

another typo damn!\[e=\sum_{n=0}^{\infty}\frac{1}{n!}<\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}\]

Answer 34

how do u know thats my upper bound?

Answer 35

the one on the right is bigger?

Answer 36

christ yet another typo
you have to add one to the number on the right, since we are starting at \(1\)

Answer 37

ok lets go back to part b

Answer 38

you showed that except for the first 1 in the sum
\[1+1+\frac{1}{2}+\frac{1}{3!}+...\]you have \(\frac{1}{n!}<\frac{1}{2^{n-1}}\)

Answer 39

is part b wrong?

Answer 40

that means that the sum of the number on the left must be less that the sum of the numbers on the right, because each number on the left (except for the first 1, which we are not counting) is smaller than each number on the right

Answer 41

no it is not wrong, i said "lets go back" because that is what we are going to use to do part c

Answer 42

oh okay:) proceed then...:)

Answer 43

we are using this fact: if \(a_n\le b_n\) for all \(n\) then
\[\sum a_n\le\sum b_n\] which should be more or less obvious right?

Answer 44

so if the sum on the left is less than the sum on the right then why is the sum on the left the upper bound?

Answer 45

it is not, the sum on the right, the one you know how to compute, is the upper bound

Answer 46

okay then that makes WAY more sense!

Answer 47

question says " Find an upper bound for e using part (b)"

Answer 48

since you know how to compute
\[\sum \frac{1}{2^n}\] this should now be easy, just don't forget to add 1 to your answer, because of the first 1 in \(1+1+\frac{1}{2}+\frac{1}{3!}+...\)

Answer 49

you want to post the question here?

Answer 50

what i did was An=(1/3^n+1) and Bn=(1/3^n) and did the lim of An/Bn and it equals 1