Question

Proofs questions.

Answer 1

For the sequence an I think it has to be an alternating series whose limit is 0. Because if an is a decreasing sequence then the whole thing would pass the Abel's test which converges and I need part c to diverge.

Answer 2

And i dont see how an could be an increasing sequence and have a limit go to 0. And then an>0which I think he means that it has to be a positive series.

Answer 3

odd thing is that if
\[a_n\] is monotonic then there is a theorem that says
\[\sum a_n b_n\] converges, so it must have something to do with monotonicity of an

Answer 4

I didnt know about that theorem.

Answer 5

What if an was the sequence 1/2 , 1/2^2, 1/3 , 1/3^2

Answer 6

That multiplies each term of a convergent series by a number less than one. That won't make it diverge.

Answer 7

http://www.math.upenn.edu/~nate/teaching/2009/spring/math_360/homework/homework_6/solutions_6.pdf

Answer 8

but that proves it converges. i want it to diverge

Answer 9

yeah i know i am trying to think of what is different

Answer 10

you only have
\[\sum b_k\] bounded so it need not converge; and in
\[b_k=(-1)^k\] but this doesn't seem to help

Answer 11

because if you alternate a sequence that goes to zero the sum will converge, so that is the wrong track

Answer 12

my teacher said this was easy too, which irks me.

Answer 13

and i figured bn would be an alternating series

Answer 14

well what do we now that is positive and goes to zero? \[a_n=\frac{1}{n}\] is a good choice

Answer 15

what if we chose the alternating harmonic series.\[a_{n} = (a^{n+1})/n\]

Answer 16

oops that a is suppose to be a -1

Answer 17

\[a _{n} = -1^{n+1}\div n\]

Answer 18

and we know that converges by the alternating series test because 1/n is monotonically decreasing, positive, and has limit 0.

Answer 19

and also it is conditionally convergent

Answer 20

and then bn = (-1)^n

Answer 21

ok here we go
i would like to claim credit, but i cannot

Answer 22

see 24.5
http://math.stanford.edu/~ksound/Math171S10/Hw3Sol_171.pdf

Answer 23

23.5?

Answer 24

any ideas animal?

Answer 25

Need to devise a way to pick up only positive or negative terms, maybe?

Answer 26

someone told me that if i had bn be an alternating series, and have an where anbn is positive then it would work.

Answer 27

Maybe\[a _{k}=1/k\]k even and \[a _{k}=1/k ^{k}\]k odd\[b _{k}=(-1)^{k}\]

Answer 28

\[k^k\] is a little overkill

Answer 29

I think that will work. The odd part of the series will get really small, really fast, and the even part will diverge.

Answer 30

More's law: If some is good, more is better, and too much is just right.

Answer 31

Thanks for letting me play; it was fun.

Answer 32

I like the sequence

\[\left\{\frac{1}{1},\frac{1}{1^2},\frac{1}{2},\frac{1}{2^2},\frac{1}{3},\frac{1}{3^2},\ldots\right\}\]

Answer 33

not that there is anything wrong with what you wrote

Answer 34

I thought about that, but it seemed not aggressive enough for me.

Answer 35

i was thinking about that sequence zarkon

Answer 36

Nice work AnimalAin.

Answer 37

Thanks. Proof even a blind squirrel finds an acorn once in a while.

Answer 38

Never discount a good sense of smell ;)

Answer 39

okay. so whose should i use?

Answer 40

doesn't matter

Answer 41

both work

Answer 42

okay. hey animal. for the summation part on C. what would that look like since I have one for even and one for odd in an

Answer 43

It looks the same as it does in the problem. When you define \[a _{k}\]just do it sort of like a piecewise function as I did. If you have to show the divergence in detail, establish a couple of other indexes and show the even terms summed under one, and the odds summed under another.

Answer 44

cool

Answer 45

Maybe like\[\sum_{k=1}^{\infty}a _{k}b _{k}=\sum_{j=1}^{\infty}a _{2j}b _{2j}+\sum_{n=1}^{\infty}a _{2n-1}b _{2n-1}\]

Answer 46

I think it's clear that one of the series is finite, and the other diverges.

Answer 47

thanks again if you check this