Limit of $a_n=(1- frac{1}{n})^n$ as $nrightarrow infty$
I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.
calculus
add a comment |
I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.
calculus
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
1
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32
add a comment |
I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.
calculus
I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.
calculus
calculus
asked Nov 25 '18 at 7:28
Wesley Strik
1,493422
1,493422
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
1
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32
add a comment |
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
1
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
1
1
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32
add a comment |
5 Answers
5
active
oldest
votes
Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.
If you are only given $(1 + 1/n)^n to e$, then
$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
add a comment |
Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$
add a comment |
Simply write:
$$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
add a comment |
You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $
add a comment |
Or you could use
$$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3012539%2flimit-of-a-n-1-frac1nn-as-n-rightarrow-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.
If you are only given $(1 + 1/n)^n to e$, then
$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
add a comment |
Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.
If you are only given $(1 + 1/n)^n to e$, then
$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
add a comment |
Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.
If you are only given $(1 + 1/n)^n to e$, then
$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$
Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.
If you are only given $(1 + 1/n)^n to e$, then
$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$
answered Nov 25 '18 at 7:49
RRL
48.9k42573
48.9k42573
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
add a comment |
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
Oh that is even nicer than the argument I came up with just now :D Thank you.
– Wesley Strik
Nov 25 '18 at 7:56
1
1
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
@WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
– RRL
Nov 25 '18 at 8:00
add a comment |
Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$
add a comment |
Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$
add a comment |
Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$
Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$
answered Nov 25 '18 at 7:54
Wesley Strik
1,493422
1,493422
add a comment |
add a comment |
Simply write:
$$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
add a comment |
Simply write:
$$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
add a comment |
Simply write:
$$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$
Simply write:
$$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$
edited Nov 25 '18 at 7:43
answered Nov 25 '18 at 7:37
Wesley Strik
1,493422
1,493422
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
add a comment |
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
1
1
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
$left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
– Yadati Kiran
Nov 25 '18 at 7:40
add a comment |
You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $
add a comment |
You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $
add a comment |
You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $
You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $
answered Nov 25 '18 at 7:49
DeepSea
70.9k54487
70.9k54487
add a comment |
add a comment |
Or you could use
$$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$
add a comment |
Or you could use
$$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$
add a comment |
Or you could use
$$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$
Or you could use
$$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$
answered Nov 25 '18 at 8:04
KM101
4,886421
4,886421
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3012539%2flimit-of-a-n-1-frac1nn-as-n-rightarrow-infty%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
$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30
haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31
It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31
1
Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32