Really quick analysis question...

is there a sequence a_n for which |a_n| doesn't converge but a_n does?

I don't think so...??

EDIT: Also, is the following a correct definition of a_n is not null

\displaystyle \exists \epsilon > 0 s.t \forall N \in \mathbb{N}, \exists n > N s.t |a_n|> \epsilon

. I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.

(edited 13 years ago)

Reply 9

PhyMath

Original post by miml

Really quick analysis question...

is there a sequence a_n for which |a_n| doesn't converge but a_n does?

I don't think so...??

EDIT: Also, is the following a correct definition of a_n is not null

Unparseable latex formula:

\displaystyle \exists \episilon > 0 s.t \forall N \in \mathbb{N}, \exists n > N s.t |a_n|> \episilon

. I don't think so, because it looks like a_n = 1/n might satisfy this definition, but I'm not sure.

I think you're right.

Time to practice some latex.

Suppose

a_n \rightarrow a

\forall \epsilon>0, \exists N \in N

such that

|a_n-a|<\epsilon ,\forall n>N

Then by the reverse triangle inequality

||a_n| - |a||< \epsilon, \forall n>N

Hence

|a_n| \rightarrow |a|

Is this right?

Reply 10

miml

Original post by PhyMath

I think you're right.

Time to practice some latex.

Suppose

a_n \rightarrow a

\forall \epsilon>0, \exists N \in N

such that

|a_n-a|<\epsilon ,\forall n>N

Then by the reverse triangle inequality

||a_n| - |a||< \epsilon, \forall n>N

Hence

|a_n| \rightarrow |a|

Is this right?

Yeah, this is the way I did it. Although admittedly I was trying to prove ||a_n|-a|<|a_n-a|, which isn't quite right.

Reply 11

username219321

I've done all the questions from Foundations/Analysis now from the booklet the MORSE society gave - I think I am gonna upload a few answers to here later but it'll take absolutely ages to latex all of them :\

Reply 12

username219321

All 89 True/False questions from section 2.1.1 of the booklet from the MORSE Society:

Spoiler

(edited 13 years ago)

Reply 13

placenta medicae talpae

Original post by matt2k8

it'll take absolutely ages to latex all of them :\

Lol, be grateful you don't have to do the ST113 programming mathematica/LaTeX assignment this Christmas! :P

Reply 14

username219321

Original post by placenta medicae talpae

Lol, be grateful you don't have to do the ST113 programming mathematica/LaTeX assignment this Christmas! :P

Be grateful I'm not a stats student or that I don't have to do loads of LaTeX'ing? :tongue:

Reply 15

Narev

Be grateful you're not latexing the booklet? :P

Reply 16

username219321

Original post by Narev

Be grateful you're not latexing the booklet? :P

Good point :smile:

Reply 17

yeohaikal

2.72 is false. let a(n)=1 for n even and a(n)=-1 for n odd, then lim inf(a) is not eq to lim sup(a) but it's sum is conditional convergent. :smile:

Reply 18

username219321

Original post by yeohaikal

2.72 is false. let a(n)=1 for n even and a(n)=-1 for n odd, then lim inf(a) is not eq to lim sup(a) but it's sum is conditional convergent. :smile:

In your example the sequence of partial sums just oscillated between -1 and 0 so the series doesn't converge - I said it was true as it's required that

(a_n) \rightarrow 0

for the series

\Sigma a_n

to converge, but if

lim \sup \ a_n \not= \lim \inf \ a_n

then

(a_n)

does not converge to any limit

(edited 13 years ago)

Reply 19

yeohaikal

Q5 is true.. by giving a strict inequality, it just gives a stronger (sufficient) definition of not null, though definitely not necessary. is that right?