How find this $lim_{ntoinfty}left(frac{1}{a_{n}+1}+frac{1}{a_{n}+2}+cdots+frac{1}{a_{n}+b_{n}}right)=x$
$begingroup$
Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
$$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$
I only know this
$$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$
But for my problem I can't. Thank you
calculus sequences-and-series analysis limits integers
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
$endgroup$
add a comment |
$begingroup$
Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
$$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$
I only know this
$$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$
But for my problem I can't. Thank you
calculus sequences-and-series analysis limits integers
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
$endgroup$
add a comment |
$begingroup$
Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
$$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$
I only know this
$$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$
But for my problem I can't. Thank you
calculus sequences-and-series analysis limits integers
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
$endgroup$
Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
$$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$
I only know this
$$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$
But for my problem I can't. Thank you
calculus sequences-and-series analysis limits integers
calculus sequences-and-series analysis limits integers
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
edited Sep 28 '14 at 15:05
Yiorgos S. Smyrlis
63.2k1384163
63.2k1384163
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
asked Jan 19 '14 at 14:41
china mathchina math
10.2k631117
10.2k631117
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that (integral comparison test)
$$
ln big(tfrac{ell-1/n}{k-1/n}big)
=int_{kn-1}^{ell n-1}frac{dx}{x}<
frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
=ln (tfrac{ell}{k})
$$
Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
$$
frac{ell_n}{k_n}tomathrm{e}^x.
$$
Then
$$
lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
frac{1}{k_nn+j}to x.
$$
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
$endgroup$
add a comment |
$begingroup$
Hint: Change the limit in a Riemann sum :-)
like here
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
$endgroup$
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
add a comment |
$begingroup$
Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
$$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
(Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
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%2f643743%2fhow-find-this-lim-n-to-infty-left-frac1a-n1-frac1a-n2-cdots%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that (integral comparison test)
$$
ln big(tfrac{ell-1/n}{k-1/n}big)
=int_{kn-1}^{ell n-1}frac{dx}{x}<
frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
=ln (tfrac{ell}{k})
$$
Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
$$
frac{ell_n}{k_n}tomathrm{e}^x.
$$
Then
$$
lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
frac{1}{k_nn+j}to x.
$$
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
$endgroup$
add a comment |
$begingroup$
Note that (integral comparison test)
$$
ln big(tfrac{ell-1/n}{k-1/n}big)
=int_{kn-1}^{ell n-1}frac{dx}{x}<
frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
=ln (tfrac{ell}{k})
$$
Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
$$
frac{ell_n}{k_n}tomathrm{e}^x.
$$
Then
$$
lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
frac{1}{k_nn+j}to x.
$$
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
$endgroup$
add a comment |
$begingroup$
Note that (integral comparison test)
$$
ln big(tfrac{ell-1/n}{k-1/n}big)
=int_{kn-1}^{ell n-1}frac{dx}{x}<
frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
=ln (tfrac{ell}{k})
$$
Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
$$
frac{ell_n}{k_n}tomathrm{e}^x.
$$
Then
$$
lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
frac{1}{k_nn+j}to x.
$$
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
$endgroup$
Note that (integral comparison test)
$$
ln big(tfrac{ell-1/n}{k-1/n}big)
=int_{kn-1}^{ell n-1}frac{dx}{x}<
frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
=ln (tfrac{ell}{k})
$$
Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
$$
frac{ell_n}{k_n}tomathrm{e}^x.
$$
Then
$$
lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
frac{1}{k_nn+j}to x.
$$
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
edited Dec 9 '18 at 17:34
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
answered Jan 19 '14 at 14:58
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
63.2k1384163
add a comment |
add a comment |
$begingroup$
Hint: Change the limit in a Riemann sum :-)
like here
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
$endgroup$
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
add a comment |
$begingroup$
Hint: Change the limit in a Riemann sum :-)
like here
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
$endgroup$
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
add a comment |
$begingroup$
Hint: Change the limit in a Riemann sum :-)
like here
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
$endgroup$
Hint: Change the limit in a Riemann sum :-)
like here
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
answered Jan 19 '14 at 14:43
Bman72Bman72
1,9221922
1,9221922
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
add a comment |
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
$begingroup$
can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
$endgroup$
– china math
Jan 19 '14 at 14:45
add a comment |
$begingroup$
Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
$$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
(Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
add a comment |
$begingroup$
Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
$$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
(Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
add a comment |
$begingroup$
Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
$$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
(Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
$$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
(Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Sep 28 '14 at 15:39
Akiva WeinbergerAkiva Weinberger
13.8k12167
13.8k12167
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%2f643743%2fhow-find-this-lim-n-to-infty-left-frac1a-n1-frac1a-n2-cdots%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
