Question

Show that it is not true that for every error 1/m there exists n such that |Sum(k=1,n,r^k-r/(1-r)|<1/m for all r in 0 < r < 1.

I know that Sum(k=1,n,r^k) = r/(1-r)-r^(n+1)/(1-r), so the above reduces to

r^(n+1) < (1-r)/m

However, I'm not sure where to go from here. Please help!

Answer 1

\[|\sum_{k=1}^nr^k-\frac{r}{1-r}|<\frac{1}{m}\]?

Answer 2

Yes

Answer 3

as you said this is the same as \[\frac{r^{n+1}}{1-r}<\frac{1}{m}\]

Answer 4

so i think the important thing here is to know how to negate the statement
"for every error 1/m there exists n such that |Sum(k=1,n,r^k-r/(1-r)|<1/m for all r in 0 < r < 1"

Answer 5

Right.

Would simply showing that n has a dependence on r be enough to disprove the statement?

Answer 6

yes that is exactly the point
because if this were true it would work for all \(r\)

Answer 7

Ok.

So would getting to \[n<\log_{r} (1-r)-\log_{r} m-1\] be enough?

Answer 8

oh you are ahead of me. maybe you can use that

Answer 9

the negation would be there exists an \(m\) such that for all \(n\) there exists an \(r\) such that \(\frac{r^{n+1}}{1-r}\geq\frac{1}{m}\)

Answer 10

so if you can find an \(r\) that would satisfy that inequality,you have proved what you need

Answer 11

that amounts to saying that \(r\) would depend on \(n\) but the proof you get to take any \(n\) and find some \(r\) that violates that condition

Answer 12

So what I need to do is set the values for r and n and then find a value for m that makes the problem statement false?

Answer 13

i would let \(m\) be given and show that for any \(n\) we can find an \(r\) that violates that condition

Answer 14

but then i am shooting from the hip, let me think for a minute

Answer 15

i mean the snappy way to say this is that the sup of \(\frac{r^{n+1}}{1-r}\) is infinity, but i don't think that is what you want. we have to be more careful

Answer 16

What if I said that as r -> 1 \[\frac{ 1-r }{ m } \rightarrow 0\]

and \[r ^{n+1} \ \rightarrow 1\] \[\forall m,n\] ?

Answer 17

ok i had to think for a minute
what you are being asked to show is that this series is convergent, but not uniformly convergent. i am not sure if you have got to that point yet, but no matter, we can solve this i think

Answer 18

we pick an \(m\) say \(m=10\) although it hardly matters.

Answer 19

Ok

Answer 20

we want to show that given any \(n\) we can find an \(r\) such that \(\frac{r^{n+1}}{1-r}\geq \frac{1}{10}\)

Answer 21

and we pick our \(r\) depending on \(n\) so we can pick \(r=1-\frac{1}{n+1}\)

Answer 22

in simple english, we are picking an \(r\) close to 1, which will make the denominator close to 0 and so the fraction \(\frac{r^{n+1}}{1-r}\) will be large, or at least larger than \(\frac{1}{10}\)

Answer 23

we get
\[\frac{r^{n+1}}{1-r}=(n+1)(1-\frac{1}{n+1})^{n+1}\] and that should be good enough, although it is now up to you to show that that is larger than \(\frac{1}{10}\)

Answer 24

How did you get \[(n+1)(1-\frac{ 1 }{ n+1})^{n+1}\]?