Question

let (an) and (bn) be two non negative sequences. Let cn=max{an,bn} for each n∈N.

Prove the following statements or give a counter example when false

(a) If (an) and (bn) converge, so does (cn).

(b)If (an)and(bn) diverge, so does(cn

c) if the series of an and the series of bn converges then so does the series of cn.

d) if the series of an and the series of bn diverges then so does the series of cn

Answer 1

I'm thinking these are both false?

Answer 2

Do u know this one @Zarkon ?

Answer 3

this is same as last question?

Answer 4

yes...i do....again what have you tried?

Answer 5

Its similar .. this is the summation of the an and bn

Answer 6

@Zarkon well I know that in order for a series to converge it must me bounded above correct?

Answer 7

the partial sums were bounded above then yes it would converge

Answer 8

Do we have to use anything with the Cauchy criterion or no?

Answer 9

you don't need to

Answer 10

Was I correct in saying these are both false at least?

Answer 11

though you might want to

Answer 12

They are both false?

Answer 13

no they are not both false

Answer 14

I posted a and b to this question earlier and two people helped me.. the questions were if (an) and (bn) both converge then (cn) converges as well..and the same for diverge.. they told me they were both true..is that in correct?

Answer 15

we are given that \[\sum_{n=1}^{\infty}a_n\text{ and }\sum_{n=1}^{\infty}b_n\]

both converge correct

Answer 16

for (c)

Answer 17

Yes correct..without the limits

Answer 18

I edited the question so u could see a and b as well

Answer 19

(a) is true...(b) is not true

Answer 20

So then c is true and d is not true?

Answer 21

c and d are also true

Answer 22

ayi yi I don't understand any of this.. I am going to lose internet connect for a little bit. Will u please help m,e when I get nback ..Id really appreciate it

Answer 23

for (c) and (d) you can use the comparison test

Answer 24

What about for a? If I just show that can is equal to bn or an and thy both convene then cn also converges?

Answer 25

you really need two cases...when (1) an->a and bn->b and a=b

2) if a>b or b>a without loss of generality you can assume a>b

Answer 26

Ok and how would I go about disproving b?

Answer 27

counter example:

\[a_n=\left\{\begin{array}{ll} 1, & n \text{ even} \\ 2, & n \text{ odd}\\ \end{array}\right.\]
\[b_n=\left\{\begin{array}{ll} 2, & n \text{ even} \\ 1, & n \text{ odd}\\ \end{array}\right.\]

\[c_n=2\]

Answer 28

do you understand?

Answer 29

Umm not exactly

Answer 30

\[\lim_{n\to \infty}a_n\] does not converge correct (ie it diverges)

Answer 31

using my \(a_n\)

Answer 32

the sequence (starting at n=1) is 2,1,2,1,2,1,2,1,2,1,2,1.....

it does not get close to one number

Answer 33

same with \(b_n\)

1,2,1,2,1,2,1,2,1,2,1,2,1,2...

Answer 34

but \(c_n\) is the max of the two sequences at any value n...that is 2

so \[\lim_{n\to \infty}c_n=2\]

Answer 35

So therefor it doesn't converge ?

Answer 36

\(\{a_n\}\) and \(\{b_n\}\) both diverge but \(\{c_n\}\) converges

Answer 37

here we are talking about sequences not series

Answer 38

ok got i get that part

Answer 39

and for part c i use the comparisson test that says if series bk converges then ak converges and since cn is the max then it must converge as well?

Answer 40

i only know the proof of comparisson test for seqences not series..is it alot different?

Answer 41

if \[\sum_{n=1}^{\infty}a_n\text{ and }\sum_{n=1}^{\infty}b_n\]

converge then

\[\sum_{n=1}^{\infty}[a_n+b_n]\] converges


and since \[c_n=max\{a_n,b_n\}\le a_n+b_n\]

we have that

\[\sum_{n=1}^{\infty}c_n\] converges by the comparison test

Answer 42

ahh thank you!

Answer 43

@zarkon so then d is exactly the same just with divergence?

Answer 44

if \[\sum_{n=1}^{\infty}a_n\] diverges and \(a_n\le c_n\) then

\[\sum_{n=1}^{\infty}c_n\] diverges by the comparison test