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Help of summation of series

Hi,


I've been studying 'The Foundations Of Mathematics' by Ian Steward and David Tall and although I'm making progress some examples I still can't do some questions even going back a few chapters. Anyways the thing that is bugging me is that I still don't know how to work with limits particularly well even though it was covered early in the book. I can work out a limit of a sequence (finite and infinite) thanks to some youtube help. However I can't calculate the sum of a series, I've tried youtube and wikipedia but I don't quite understand whats going on. Anyways the part of question I've been trying to crack is:

For real numbers a, r and natural number n
let sn= a + ar + ... arn Show that rsn sn= a(rn+1-1)
And deduce that
| sn a1r\frac{a}{1-r} | = | rn+11r\frac{r ^{n+1}}{1-r} |
For |r| < 1 deduce that sn -> a/(1-r) as n -> infinity.

The rsn snbit it easy enough. The bit after that I got | arn+11r\frac{ar ^{n+1}}{1-r} | and I’m not sure where I made a mistake but I’m sure its minor so I’m not too bothered about that. But the last bit I know that as n approaches infinity arn will approach 0. Also if r = 0 its easy to see from the sequence and formulae we are showing that the series will equal a. However I don't know where to go from here. Can someone please point out where to go next, I would really appreciate that. Although I need help with 2 parts of this question it’s really the later part that’s bothering me cause I can’t work out infinite series L.
Reply 1
Firstly, I agree with you that it's
Unparseable latex formula:

\dfrac{ar^{n+1}}{1-r}}

, not
Unparseable latex formula:

\dfrac{r^{n+1}}{1-r}}

.

For the later bit, it's wanting you to show that
Unparseable latex formula:

\dfrac{r^{n+1}}{1-r}} \to 0

as nn \to \infty.
Reply 2
Original post by DFranklin
Firstly, I agree with you that it's
Unparseable latex formula:

\dfrac{ar^{n+1}}{1-r}}

, not
Unparseable latex formula:

\dfrac{r^{n+1}}{1-r}}

.


For the later bit, it's wanting you to show that
Unparseable latex formula:

\dfrac{r^{n+1}}{1-r}} \to 0

as nn \to \infty.


Oh I'm glad you agree with me, perhaps a typo? I tried that like 100 times! Ah yes I've just had another attempt and got to sa1r=0 |s_\infty - \frac{a}{1-r}|= 0 can I just take the a1r \frac{a}{1-r} over, I couldn't work out if I could do that with the mod around it?
Original post by henryhenry
...


If |a| = 0 , then a=0. I.e. you can just drop the modulus signs.
(edited 11 years ago)
Reply 4
Original post by ghostwalker
If |a| = 0 , then a=0. I.e. you can just drop the modulus signs.



Ah yes, keep missing small things :frown:

Thanks for the help :biggrin:

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