Show that $(1+frac{1}{n})^n=sum_{k=0}^{n}frac{1}{k!}Rightarrow...
$begingroup$
I write out the left term expression and get
$$
sum_{k=0}^{n}binom{n}{k}bigg(frac{1}{n}bigg)^k
$$
If I could Show that the k-th term of both sequences is equal I would be done.
I.e what I want to show is
$$binom{n}{k}bigg(frac{1}{n}bigg)^k=frac{ncdotldotsbig((n-k)+1big)}{k!}bigg(frac{1}{n}bigg)^koverset{!}{=}frac{1}{k!}$$
The approach must be wrong, because the above statement would be equivalent to say that
$$frac{ncdotldotsbig(n-(k-1)big)}{n^k}=1$$
which is not true.
How can I solve the Problem? I also want to ask if I could Show the left side of the implication in the title, why is the Right side true?
real-analysis sequences-and-series
$endgroup$
add a comment |
$begingroup$
I write out the left term expression and get
$$
sum_{k=0}^{n}binom{n}{k}bigg(frac{1}{n}bigg)^k
$$
If I could Show that the k-th term of both sequences is equal I would be done.
I.e what I want to show is
$$binom{n}{k}bigg(frac{1}{n}bigg)^k=frac{ncdotldotsbig((n-k)+1big)}{k!}bigg(frac{1}{n}bigg)^koverset{!}{=}frac{1}{k!}$$
The approach must be wrong, because the above statement would be equivalent to say that
$$frac{ncdotldotsbig(n-(k-1)big)}{n^k}=1$$
which is not true.
How can I solve the Problem? I also want to ask if I could Show the left side of the implication in the title, why is the Right side true?
real-analysis sequences-and-series
$endgroup$
1
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19
add a comment |
$begingroup$
I write out the left term expression and get
$$
sum_{k=0}^{n}binom{n}{k}bigg(frac{1}{n}bigg)^k
$$
If I could Show that the k-th term of both sequences is equal I would be done.
I.e what I want to show is
$$binom{n}{k}bigg(frac{1}{n}bigg)^k=frac{ncdotldotsbig((n-k)+1big)}{k!}bigg(frac{1}{n}bigg)^koverset{!}{=}frac{1}{k!}$$
The approach must be wrong, because the above statement would be equivalent to say that
$$frac{ncdotldotsbig(n-(k-1)big)}{n^k}=1$$
which is not true.
How can I solve the Problem? I also want to ask if I could Show the left side of the implication in the title, why is the Right side true?
real-analysis sequences-and-series
$endgroup$
I write out the left term expression and get
$$
sum_{k=0}^{n}binom{n}{k}bigg(frac{1}{n}bigg)^k
$$
If I could Show that the k-th term of both sequences is equal I would be done.
I.e what I want to show is
$$binom{n}{k}bigg(frac{1}{n}bigg)^k=frac{ncdotldotsbig((n-k)+1big)}{k!}bigg(frac{1}{n}bigg)^koverset{!}{=}frac{1}{k!}$$
The approach must be wrong, because the above statement would be equivalent to say that
$$frac{ncdotldotsbig(n-(k-1)big)}{n^k}=1$$
which is not true.
How can I solve the Problem? I also want to ask if I could Show the left side of the implication in the title, why is the Right side true?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 29 at 19:35


Daniele Tampieri
2,3972922
2,3972922
asked Jan 23 at 15:28
RM777RM777
38312
38312
1
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19
add a comment |
1
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19
1
1
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The condition $a_n=b_n$ for all $n$ is sufficient but not necessary for
$$
lim_{nto infty}a_n=lim_{ntoinfty}b_n.
$$
Note that
$$
lim_{ntoinfty}(1+frac{1}{n})^n=sum_{k=0}^inftyfrac{1}{k!}=:etag{1}
$$
But
$
(1+frac{1}{n})^n=sum_{k=0}^nfrac{1}{k!}
$
is not even true for $n=2$.
Here is a sketch for one way to show (1).
Let
$$
s_n=sum_{k=0}^nfrac{1}{k!},quad t_n=(1+frac{1}{n})^n.
$$
Step 1. Use the binomial theorem to show that $t_nle s_n$. (Exercise!) It then follows that
$$
limsup_{ntoinfty}t_nle e.tag{2}
$$
Step 2. If $nge m$, (applying the binomial theorem again)
$$
t_nge 1+1+frac{1}{2!}(1-frac{1}{n})+cdots+frac{1}{m!}(1-frac1n)cdots(1-frac{m-1}{n}).
$$
It follows that for any fixed $m$,
$$
s_mleliminf_{ntoinfty} t_n.
$$
Letting $mtoinfty$, one gets
$$
eleliminf_{ntoinfty} t_n.tag{3}
$$
You get the desired result by combining (2) and (3).
$endgroup$
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%2f3084616%2fshow-that-1-frac1nn-sum-k-0n-frac1k-rightarrow-lim-n-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The condition $a_n=b_n$ for all $n$ is sufficient but not necessary for
$$
lim_{nto infty}a_n=lim_{ntoinfty}b_n.
$$
Note that
$$
lim_{ntoinfty}(1+frac{1}{n})^n=sum_{k=0}^inftyfrac{1}{k!}=:etag{1}
$$
But
$
(1+frac{1}{n})^n=sum_{k=0}^nfrac{1}{k!}
$
is not even true for $n=2$.
Here is a sketch for one way to show (1).
Let
$$
s_n=sum_{k=0}^nfrac{1}{k!},quad t_n=(1+frac{1}{n})^n.
$$
Step 1. Use the binomial theorem to show that $t_nle s_n$. (Exercise!) It then follows that
$$
limsup_{ntoinfty}t_nle e.tag{2}
$$
Step 2. If $nge m$, (applying the binomial theorem again)
$$
t_nge 1+1+frac{1}{2!}(1-frac{1}{n})+cdots+frac{1}{m!}(1-frac1n)cdots(1-frac{m-1}{n}).
$$
It follows that for any fixed $m$,
$$
s_mleliminf_{ntoinfty} t_n.
$$
Letting $mtoinfty$, one gets
$$
eleliminf_{ntoinfty} t_n.tag{3}
$$
You get the desired result by combining (2) and (3).
$endgroup$
add a comment |
$begingroup$
The condition $a_n=b_n$ for all $n$ is sufficient but not necessary for
$$
lim_{nto infty}a_n=lim_{ntoinfty}b_n.
$$
Note that
$$
lim_{ntoinfty}(1+frac{1}{n})^n=sum_{k=0}^inftyfrac{1}{k!}=:etag{1}
$$
But
$
(1+frac{1}{n})^n=sum_{k=0}^nfrac{1}{k!}
$
is not even true for $n=2$.
Here is a sketch for one way to show (1).
Let
$$
s_n=sum_{k=0}^nfrac{1}{k!},quad t_n=(1+frac{1}{n})^n.
$$
Step 1. Use the binomial theorem to show that $t_nle s_n$. (Exercise!) It then follows that
$$
limsup_{ntoinfty}t_nle e.tag{2}
$$
Step 2. If $nge m$, (applying the binomial theorem again)
$$
t_nge 1+1+frac{1}{2!}(1-frac{1}{n})+cdots+frac{1}{m!}(1-frac1n)cdots(1-frac{m-1}{n}).
$$
It follows that for any fixed $m$,
$$
s_mleliminf_{ntoinfty} t_n.
$$
Letting $mtoinfty$, one gets
$$
eleliminf_{ntoinfty} t_n.tag{3}
$$
You get the desired result by combining (2) and (3).
$endgroup$
add a comment |
$begingroup$
The condition $a_n=b_n$ for all $n$ is sufficient but not necessary for
$$
lim_{nto infty}a_n=lim_{ntoinfty}b_n.
$$
Note that
$$
lim_{ntoinfty}(1+frac{1}{n})^n=sum_{k=0}^inftyfrac{1}{k!}=:etag{1}
$$
But
$
(1+frac{1}{n})^n=sum_{k=0}^nfrac{1}{k!}
$
is not even true for $n=2$.
Here is a sketch for one way to show (1).
Let
$$
s_n=sum_{k=0}^nfrac{1}{k!},quad t_n=(1+frac{1}{n})^n.
$$
Step 1. Use the binomial theorem to show that $t_nle s_n$. (Exercise!) It then follows that
$$
limsup_{ntoinfty}t_nle e.tag{2}
$$
Step 2. If $nge m$, (applying the binomial theorem again)
$$
t_nge 1+1+frac{1}{2!}(1-frac{1}{n})+cdots+frac{1}{m!}(1-frac1n)cdots(1-frac{m-1}{n}).
$$
It follows that for any fixed $m$,
$$
s_mleliminf_{ntoinfty} t_n.
$$
Letting $mtoinfty$, one gets
$$
eleliminf_{ntoinfty} t_n.tag{3}
$$
You get the desired result by combining (2) and (3).
$endgroup$
The condition $a_n=b_n$ for all $n$ is sufficient but not necessary for
$$
lim_{nto infty}a_n=lim_{ntoinfty}b_n.
$$
Note that
$$
lim_{ntoinfty}(1+frac{1}{n})^n=sum_{k=0}^inftyfrac{1}{k!}=:etag{1}
$$
But
$
(1+frac{1}{n})^n=sum_{k=0}^nfrac{1}{k!}
$
is not even true for $n=2$.
Here is a sketch for one way to show (1).
Let
$$
s_n=sum_{k=0}^nfrac{1}{k!},quad t_n=(1+frac{1}{n})^n.
$$
Step 1. Use the binomial theorem to show that $t_nle s_n$. (Exercise!) It then follows that
$$
limsup_{ntoinfty}t_nle e.tag{2}
$$
Step 2. If $nge m$, (applying the binomial theorem again)
$$
t_nge 1+1+frac{1}{2!}(1-frac{1}{n})+cdots+frac{1}{m!}(1-frac1n)cdots(1-frac{m-1}{n}).
$$
It follows that for any fixed $m$,
$$
s_mleliminf_{ntoinfty} t_n.
$$
Letting $mtoinfty$, one gets
$$
eleliminf_{ntoinfty} t_n.tag{3}
$$
You get the desired result by combining (2) and (3).
answered Jan 29 at 18:16
user587192
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%2f3084616%2fshow-that-1-frac1nn-sum-k-0n-frac1k-rightarrow-lim-n-right%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
1
$begingroup$
You might cannot prove that $(1+1/n)^n = sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly.
$endgroup$
– xbh
Jan 23 at 15:36
$begingroup$
It's not true : $(1+frac{1}{2})^2=frac{9}{4}$ while $1+frac{1}{1!}+frac{1}{2!}=2,5$.
$endgroup$
– Ayoub
Jan 23 at 15:37
$begingroup$
Since $pimplies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$.
$endgroup$
– user587192
Jan 29 at 18:19