Seeing that $lim_{x to infty} sum (-x)^n/n! = 0$
$begingroup$
Is there any way to see directly from the power series that $$lim_{x to infty} sum_{n=0}^infty frac{(-x)^n}{n!} = 0$$? I realize that $displaystyle{lim_{x to infty} e^{-x} = 0}$. That's not what I'm asking.
calculus sequences-and-series limits power-series
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
$endgroup$
add a comment |
$begingroup$
Is there any way to see directly from the power series that $$lim_{x to infty} sum_{n=0}^infty frac{(-x)^n}{n!} = 0$$? I realize that $displaystyle{lim_{x to infty} e^{-x} = 0}$. That's not what I'm asking.
calculus sequences-and-series limits power-series
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
$endgroup$
5
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40
add a comment |
$begingroup$
Is there any way to see directly from the power series that $$lim_{x to infty} sum_{n=0}^infty frac{(-x)^n}{n!} = 0$$? I realize that $displaystyle{lim_{x to infty} e^{-x} = 0}$. That's not what I'm asking.
calculus sequences-and-series limits power-series
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
$endgroup$
Is there any way to see directly from the power series that $$lim_{x to infty} sum_{n=0}^infty frac{(-x)^n}{n!} = 0$$? I realize that $displaystyle{lim_{x to infty} e^{-x} = 0}$. That's not what I'm asking.
calculus sequences-and-series limits power-series
calculus sequences-and-series limits power-series
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
edited Dec 6 '18 at 4:17
zoidberg
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
asked Dec 6 '18 at 4:02
zoidbergzoidberg
1,065113
1,065113
5
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40
add a comment |
5
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40
5
5
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just working with series and the Cauchy product we have
$$sum_{n=0}^infty frac{(-x)^n}{n!}sum_{n=0}^infty frac{x^n}{n!} = sum_{n=0}^inftysum_{k=0}^n frac{x^k}{k!}frac{(-x)^{n-k}}{(n-k)!} = sum_{n=0}^inftyfrac{(x+(-x))^n}{n!} = 1$$
and it is easy to show that as $x to infty$
$$sum_{n=0}^infty frac{x^n}{n!} to +infty$$
Hence, ...
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
$endgroup$
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%2f3028019%2fseeing-that-lim-x-to-infty-sum-xn-n-0%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
$begingroup$
Just working with series and the Cauchy product we have
$$sum_{n=0}^infty frac{(-x)^n}{n!}sum_{n=0}^infty frac{x^n}{n!} = sum_{n=0}^inftysum_{k=0}^n frac{x^k}{k!}frac{(-x)^{n-k}}{(n-k)!} = sum_{n=0}^inftyfrac{(x+(-x))^n}{n!} = 1$$
and it is easy to show that as $x to infty$
$$sum_{n=0}^infty frac{x^n}{n!} to +infty$$
Hence, ...
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
$endgroup$
add a comment |
$begingroup$
Just working with series and the Cauchy product we have
$$sum_{n=0}^infty frac{(-x)^n}{n!}sum_{n=0}^infty frac{x^n}{n!} = sum_{n=0}^inftysum_{k=0}^n frac{x^k}{k!}frac{(-x)^{n-k}}{(n-k)!} = sum_{n=0}^inftyfrac{(x+(-x))^n}{n!} = 1$$
and it is easy to show that as $x to infty$
$$sum_{n=0}^infty frac{x^n}{n!} to +infty$$
Hence, ...
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
$endgroup$
add a comment |
$begingroup$
Just working with series and the Cauchy product we have
$$sum_{n=0}^infty frac{(-x)^n}{n!}sum_{n=0}^infty frac{x^n}{n!} = sum_{n=0}^inftysum_{k=0}^n frac{x^k}{k!}frac{(-x)^{n-k}}{(n-k)!} = sum_{n=0}^inftyfrac{(x+(-x))^n}{n!} = 1$$
and it is easy to show that as $x to infty$
$$sum_{n=0}^infty frac{x^n}{n!} to +infty$$
Hence, ...
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
$endgroup$
Just working with series and the Cauchy product we have
$$sum_{n=0}^infty frac{(-x)^n}{n!}sum_{n=0}^infty frac{x^n}{n!} = sum_{n=0}^inftysum_{k=0}^n frac{x^k}{k!}frac{(-x)^{n-k}}{(n-k)!} = sum_{n=0}^inftyfrac{(x+(-x))^n}{n!} = 1$$
and it is easy to show that as $x to infty$
$$sum_{n=0}^infty frac{x^n}{n!} to +infty$$
Hence, ...
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
answered Dec 6 '18 at 4:31
RRLRRL
50.2k42573
50.2k42573
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%2f3028019%2fseeing-that-lim-x-to-infty-sum-xn-n-0%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
5
$begingroup$
What about showing $$sum_{n ge 0} frac{(-x)^{n}}{n!} = frac{1}{sum_{n ge 0} frac{x^{n}}{n!}}$$ and then taking limits of both sides.
$endgroup$
– Mattos
Dec 6 '18 at 4:29
$begingroup$
Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left.
$endgroup$
– zoidberg
Dec 6 '18 at 4:32
$begingroup$
Please replace $infty$ with $color{red}{+}infty$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 15:40