How to prove that $\frac{1}{9}= \sum_{i = 1}^{ \infty } \frac{1}{10^i}$

Question

I am new here and I am eager to find out how to prove this: $\frac{1}{9}= \sum_{i = 1}^{ \infty } \frac{1}{10^i}$ Is induction a method in order to prove that? Even though, to my knowledge, induction only works for finite numbers. I really want to know the answer. Thanks a lot.

Why not just use the formula for a sum of a geometric series? — Brenton, Apr 11 '16 at 18:58
This is not a proof: $$\frac9{10}+\frac9{100}+\frac9{1000}+\cdots=0.9+0.09+0.009+\cdots=0.999\cdots=1$$ — user296113, Apr 11 '16 at 18:59
It is a geometric series https://en.wikipedia.org/wiki/Geometric_series#Formula — Ángel Mario Gallegos, Apr 11 '16 at 19:00
I have a math theory which includes sums and I already have direct proof for a part of it. — OpenHax, Apr 11 '16 at 19:00
I just want to understand how it would be proven for future work. — OpenHax, Apr 11 '16 at 19:01
@OpenHax What are you referring to about stealing? This is a very well-known result that (as far as I know) dates back to the ancient Greeks. — , Apr 11 '16 at 19:03
Guys I am 14 and I never been taught geometric series or something of that sort. I figured out a interesting thing and I made some formulas which would predict the outcome. I already have direct proof for a part of my theory. But I just want to extend my knowledge. I am not a mathematician I usually do programming I just wanted to understand more. Thanks for those who intended to help. — OpenHax, Apr 11 '16 at 19:04
T Bongers I am not talking about the formula that I wrote above. — OpenHax, Apr 11 '16 at 19:05
"I mean everyone can steal adapt and post in their own name. ", no offense meant, but at that level almost everything is prior art and someone would need real lots of money to bribe the authorities for ignoring this fact. — Gyro Gearloose, Apr 11 '16 at 19:05
@OpenHax It's great that you applied yourself and found another formula. You should definitely keep doing that. However, at this level everything is already known, so either your formula is wrong or it's a known one. I encourage you to post it so that you can learn further. — rubik, Apr 11 '16 at 19:08
rubik thanks for being so kind and honest to a newbie in mathematics like me — OpenHax, Apr 11 '16 at 19:20
My formula it's not at this level for sure. It's more complex and more elegant and yes it is possible that it has been found by someone else but I searched and didn't find anything. I can't wait to show my work to you. I first want to 100% prove it. As I said, I already have some direct proof for a part of it and now I have to clear some things up and put it next to a nice formulation like a nice formula deserves. — OpenHax, Apr 11 '16 at 19:23

score 3 · Accepted Answer · answered Apr 11 '16 at 19:01

3

$A=\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...$

$10A=1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...$

$10A-A=1$

$A=\frac{1}{9}$

So the sum really equals $\frac{1}{9}$.

answered Apr 11 '16 at 19:01

The beauty is in the simplicity +1 – daniels_pa Apr 11 '16 at 19:22
Thanks for explaining. As @daniels_pa said, the beauty is in the simplicity and I really appreciate that. Thanks. – OpenHax Apr 11 '16 at 19:26
This isn't a proof. $1{+}2{+}4{+}\ldots=({-}1{+}2){+}({-}2{+}4){+}({-}4{+}8){+}\ldots={-}1{+}(2{-}2){+}(4{-}4){+}\ldots={-}1$ – user326572 Apr 11 '16 at 19:37
Then can you please provide proof? – OpenHax Apr 11 '16 at 20:03
I would appreciate it. Thanks. – OpenHax Apr 11 '16 at 20:04
@OpenHax Nope. I think the accepted answer of the question which this is a duplicate of does explain it pretty well. – user326572 Apr 12 '16 at 00:25

score 0 · Answer 2 · answered Apr 11 '16 at 19:04

There are much easier ways of doing this, but if you really want to use induction:

You can show the following finite case:

$\frac{1}{9}$ written out to $k$ decimal places is equal to $\sum_{i = 1}^k \frac{1}{10^i}$ for all $k \geq 1$

If you prove this, for the infinite case you know that this will be true for $k \rightarrow \infty$

score 0 · Answer 3 · answered Apr 11 '16 at 19:05

$$S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 1 }{ { 10 }^{ 1 } } +\frac { 1 }{ { 10 }^{ 2 } } +\frac { 1 }{ { 10 }^{ 3 } } +\cdots \\ \frac { S }{ 10 } =\frac { 1 }{ { 10 }^{ 1 } } +\frac { 1 }{ { 10 }^{ 2 } } +\frac { 1 }{ { 10 }^{ 3 } } +\frac { 1 }{ { 10 }^{ 4 } } +\cdots =\sum _{ i=1 }^{ \infty }{ \frac { 1 }{ { 10 }^{ i } } } \\ S-\frac { S }{ 10 } =\frac { 1 }{ { 10 }^{ 0 } } =1\\ \frac { 9S }{ 10 } =1\\ \frac { S }{ 10 } =\frac { 1 }{ 9 } $$

How to prove that $\frac{1}{9}= \sum_{i = 1}^{ \infty } \frac{1}{10^i}$

3 Answers3