Integral that is continuous and looks like it converges to a geometric seriesTesting if a geometric series...
Was Dennis Ritchie being too modest in this quote about C and Pascal?
Trouble removing package using Yum on CentOS7
Are there moral objections to a life motivated purely by money? How to sway a person from this lifestyle?
Which big number is bigger?
How do I reattach a shelf to the wall when it ripped out of the wall?
What does "function" actually mean in music?
What is the term for a person whose job is to place products on shelves in stores?
Is Electric Central Heating worth it if using Solar Panels?
Mistake in years of experience in resume?
What is this word supposed to be?
Retract an already submitted recommendation letter (written for an undergrad student)
Will I lose my paid in full property
A strange hotel
Injection into a proper class and choice without regularity
Where was the County of Thurn und Taxis located?
Should the Product Owner dictate what info the UI needs to display?
Can a level 2 Warlock take one level in rogue, then continue advancing as a warlock?
Is there a better way to say "see someone's dreams"?
Could moose/elk survive in the Amazon forest?
Drawing a german abacus as in the books of Adam Ries
Check if a string is entirely made of the same substring
Negative Resistance
Is Diceware more secure than a long passphrase?
I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?
Integral that is continuous and looks like it converges to a geometric series
Testing if a geometric series converges by taking limit to infinitySummation of arithmetic-geometric series of higher orderGeometric series with polynomial exponentHow to Recognize a Geometric SeriesShowing an integral equality with series over the integersDiscontinuity of a series of continuous functionsReasons why a Series ConvergesSum of infinite geometric series with two terms in summationUsing geometric series for computing IntegralsLimit of geometric series sum when $r = 1$
$begingroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
add a comment |
$begingroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
add a comment |
$begingroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
1026
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
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%2f3202627%2fintegral-that-is-continuous-and-looks-like-it-converges-to-a-geometric-series%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
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
add a comment |
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
add a comment |
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
43k23482
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
add a comment |
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
4 hours ago
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%2f3202627%2fintegral-that-is-continuous-and-looks-like-it-converges-to-a-geometric-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
