Mixing
Rational as series?
$begingroup$
I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...
sequences-and-series reference-request rational-numbers geometric-series
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
$endgroup$
add a comment |
$begingroup$
I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...
sequences-and-series reference-request rational-numbers geometric-series
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
$endgroup$
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03
add a comment |
$begingroup$
I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...
sequences-and-series reference-request rational-numbers geometric-series
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
$endgroup$
I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...
sequences-and-series reference-request rational-numbers geometric-series
sequences-and-series reference-request rational-numbers geometric-series
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
edited Feb 2 at 6:35
Mahiyan Intesir Prethu
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
asked Feb 2 at 5:40
Mahiyan Intesir PrethuMahiyan Intesir Prethu
64
64
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03
add a comment |
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.
I don't know who showed this first, but this is pretty straightforward.
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
$endgroup$
add a comment |
$begingroup$
The formula you wrote is not entirely correct. What you have is
$$tag1
frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
$$
This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.
As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097056%2frational-as-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.
I don't know who showed this first, but this is pretty straightforward.
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
$endgroup$
add a comment |
$begingroup$
You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.
I don't know who showed this first, but this is pretty straightforward.
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
$endgroup$
add a comment |
$begingroup$
You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.
I don't know who showed this first, but this is pretty straightforward.
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
$endgroup$
You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.
I don't know who showed this first, but this is pretty straightforward.
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
answered Feb 2 at 6:16
Clement C.Clement C.
51.1k34093
51.1k34093
add a comment |
add a comment |
$begingroup$
The formula you wrote is not entirely correct. What you have is
$$tag1
frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
$$
This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.
As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
The formula you wrote is not entirely correct. What you have is
$$tag1
frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
$$
This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.
As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
The formula you wrote is not entirely correct. What you have is
$$tag1
frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
$$
This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.
As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
The formula you wrote is not entirely correct. What you have is
$$tag1
frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
$$
This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.
As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 14:50
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097056%2frational-as-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45
$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50
$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55
$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03